datavorous/entrance-exam-dataset huggingface dataset:
In the circuit shown below, is working as a 8 V dc regulated voltage source. When 12 V is used as an input, the power dissipated (in mW ) in each diode is (Considering both zener diodes are identical)

Options:

The correct answer is:
40

Solution:
i=12-8200+200A=4400=10-2 A
Power loss in each diode=410-2 W=40 mW

Topic: Physics, Semiconductors, JEE Main, JEE Main 2020 (09 Jan Shift 2)
In the circuit shown below (on the left) the resistance and the emf source are both variable. The graph of seven readings of the voltmeter and the ammeter (V and I, respectively) for different setting of resistance and the emf, taken at equal intervals of time $\Delta t$, are shown (on the right) by the dots connected by the curve EFGH. Consider the interval resistance of the battery to be negligible and the voltmeter and ammeter to be ideal devices. Take $R_{0}=V_{0} I_{0}$
Then the plot of the resistance as a function of time corresponding to the curve EFGH is given by

Options:
A
B
C
D

The correct answer is:

Solution:
From $\mathrm{E} \rightarrow \mathrm{F}$
Slope is constant
$V=\mid R$
Thus $R$ must be constant
$\mathrm{R}_{0}=\mathrm{V}_{0} / \mathrm{I}_{0}$
Frorm $\mathrm{F} \rightarrow \mathrm{G}$
$\mathrm{V}=$ constant
$\mathrm{I}$ is increase thus $\mathrm{R}$ must be decrease.
at $G, R=\frac{V_{0}}{2 I_{0}}=\frac{R_{0}}{2}$
From $\mathbf{G} \rightarrow \mathbf{H}$
$\mathrm{I}=$ constant
$\mathrm{V}=$ increase, thus $\mathrm{R}$ must be increase
$\mathrm{R}_{\mathrm{H}}=\frac{2 \mathrm{~V}_{0}}{2 \mathrm{l}_{0}}=\mathrm{R}_{0}$

Topic: Physics, Current Electricity, JEE Main
In the circuit shown below, the ac source has voltage $V=20 \cos (\omega t)$ volts with $\omega=2000 \mathrm{rad} / \mathrm{sec}$. the amplitude of the current will be nearest to

Options:
A $2 A$
B $3.3A$
C $2 / \sqrt{5} A$
D $2.8 A$

The correct answer is:
$2 A$

Solution:
$\begin{aligned} & R=6+4=10 \Omega \\ & X_L=\omega L=2000 \times 5 \times 10^{-3}=10 \Omega \\ & X_C=\frac{1}{\omega C}=\frac{1}{2000 \times 50 \times 10^{-6}}=10 \Omega \\ & \therefore Z=\sqrt{R^2+\left(X_L-X_C\right)^2}=10 \Omega \\ & \text { Amplitude of current }=i_0=\frac{V_0}{Z}=\frac{20}{10}=2 \mathrm{~A}\end{aligned}$

Topic: Physics, Alternating Current, JEE Main
In the circuit shown below, the key $\mathrm{K}$ is closed at $\mathrm{t}=0$. The current through the battery is

Options:
A
$\frac{\mathrm{VR}_1 \mathrm{R}_2}{\sqrt{\mathrm{R}_1^2+\mathrm{R}_2^2}}$ at $\mathrm{t}=0$ and $\frac{\mathrm{V}}{\mathrm{R}_2}$ at $\mathrm{t}=\infty$

B
$\frac{V}{R_2}$ at $t=0$ and $\frac{V\left(R_1+R_2\right)}{R_1 R_2}$ at $t=\infty$

C
$\frac{V}{R_2}$ at $t=0$ and $\frac{V R_1 R_2}{\sqrt{R_1^2+R_2^2}}$ at $t=\infty$

D
$\frac{V\left(R_1+R_2\right)}{R_1 R_2}$ at $t=0$ and $\frac{V}{R_2}$ at $t=\infty$

The correct answer is:
$\frac{V}{R_2}$ at $t=0$ and $\frac{V\left(R_1+R_2\right)}{R_1 R_2}$ at $t=\infty$

Solution:
At $\mathrm{t}=0$, inductor behaves like an infinite resistance
So at $t=0, i=\frac{V}{R_2}$ and at $t=\infty$, inductor behaves like a conducting wire
$$
\mathrm{i}=\frac{\mathrm{V}}{\mathrm{R}_{\text {eq }}}=\frac{\mathrm{V}\left(\mathrm{R}_1+\mathrm{R}_2\right)}{\mathrm{R}_1 \mathrm{R}_2}
$$

Topic: Physics, Electromagnetic Induction, JEE Main, JEE Main 2010
In the circuit shown below, The reading of the voltmeter $V$ is

Options:
A $12 \mathrm{~V}$
B $8 \mathrm{~V}$
C $20 \mathrm{~V}$
D $16 \mathrm{~V}$

The correct answer is:
$12 \mathrm{~V}$

Solution:
In the following circuit potential difference between
$C$ and $A$ is $V_C-V_A=1 \times 4=4$ ....(i)
$C$ and $B$ is $V_C-V_B=1 \times 16=16$ ...(ii)
On solving equations (i) and (ii) we get
$V_A-V_B=12 \mathrm{~V} \text {. }$

Topic: Physics, Current Electricity, JEE Main
In the circuit shown below, the voltmeter is of large resistance. The E.M.F. of the cell is ε. The reading of the voltmeter is

Options:
A zero
B ε10
C ε5
D ε2

The correct answer is:
ε5

Solution:
From the figure, it is clear that
VA â VB =Îµ4+6 Ã 4=4Îµ10
VA â VD =Îµ6+4 Ã 6 =6Îµ10
VB â VD=Îµ5

Topic: Physics, Current Electricity, JEE Main
In the circuit shown, charge on the 5μF capacitor is :

Options:
A 18.00μC
B 10.90μC
C 16.36μC
D 5.45μC

The correct answer is:
5.45μC

Solution:
VB=6 VC=x VD=6

-91+92+93=0
-2(6-x)+4(x-6)+5(x-0)=0
-12+2 x+4 x-24+5 x=0
11x=36⇒x=3611;q3=5×3611=18011μC

Topic: Physics, Capacitance, JEE Main, JEE Main 2020 (05 Sep Shift 2)
In the circuit shown, current (in A) through $50 \mathrm{~V}$ and $30 \mathrm{~V}$ batteries are, respectively:

Options:
A
$2.5$ and 3

B
$3.5$ and 2

C
$4.5$ and 1

D
3 and $2.5$

The correct answer is:
$2.5$ and 3

Solution:
Current through $50 \mathrm{~V}$ and $30 \mathrm{~V}$ batteries are respectively $2.5 \mathrm{~A}$ and $3 \mathrm{~A}$.

Topic: Physics, Current Electricity, JEE Main, JEE Main 2014 (11 Apr Online)
In the circuit shown, current (in A) through the 50 V and 30 V batteries are, respectively

Options:
A 2.5 and 3
B 4.5 and 1
C 5.5 and 2
D 3.5 and 2

The correct answer is:
4.5 and 1

Solution:
Applying Kirchoff's loop rule in ABCDA
50 â 20I1 = 0
â´I1=2Â·5 A
Applying Kirchoff's loop rule in EGHFE

-10I2+30=0
â´I2=3 A
Applying Kirchoff's loop rule in CEFDC
-5I-I1-10I2-5I-I1+20I1=0
- 1 0 I - 2 Â· 5 - 3 0 + 2 0 Ã 2 Â· 5 = 0
- 1 0 I + 2 5 - 3 0 + 5 0 = 0
I=4Â·5 A
I-I1-I2=4Â·5-2Â·5-3=-1

4.5 A and 1 A

Topic: Physics, Current Electricity, JEE Main
In the circuit shown, current measured by ammeter is x ampere. Find x.

Options:

The correct answer is:
03.00

Solution:
Let current = x ampere
V2+V3=40 V (L & C out of phase)
Voltage across 10 Ω=10x
Using phase addition of voltage,
(10x)2+(40)2=(50)2
⇒x=3

Topic: Physics, Alternating Current, JEE Main
In the circuit shown, each resistor is of resistance R. The equivalent resistance between terminals A and B is :

Options:
A 2R
B 1.3 R
C 1.5R
D 15 R

The correct answer is:
1.5R

Solution:
[balanced wheat-stone bridge] then equivalent resistance, =3R2=1.5 R

Topic: Physics, Current Electricity, JEE Main
In the circuit shown, each resistor is of resistance R. The equivalent resistance between the terminals A and B is

Options:
A 2 R
B 1.3 R
C 1.5âR
D 15 R

The correct answer is:
1.5 R

Solution:
[The given circuit is actually a balanced wheat-stone bridge]

Hence, equivalent resistance=3R2=1.5R

Topic: Physics, Current Electricity, JEE Main
In the circuit shown here $C_1=6 \mu F, C_2=3 \mu F$ and battery $B=20 \mathrm{~V}$. The switch $S_1$ is first closed. It is then opened and afterwards $S_2$ is closed. What is the charge finally on $C_2$

Options:
A $120 \mu \mathrm{C}$
B $80 \mu \mathrm{C}$
C $40 \mu \mathrm{C}$
D $20 \mu \mathrm{C}$

The correct answer is:
$40 \mu \mathrm{C}$

Solution:
Common potential $V=\frac{6 \times 20+3 \times 0}{(6+3)}=\frac{120}{9} \mathrm{Voft}$ So, charge on $3 \mu F$ capacitor
$Q_2=3 \times 10^{-6} \times \frac{120}{a}=40 \mu C$

Topic: Physics, Capacitance, JEE Main
In the circuit shown here, the point 'C' is kept connected to point A '' till the current flowing through the circuit becomes constant. Afterward, suddenly, point 'C' is disconnected from point A '' and connected to point 'B' at time t=0. Ratio of the voltage across resistance and the inductor at t=LR will be equal to :

Options:
A e 1 - e
B 1
C - 1
D 1 - e e

The correct answer is:
- 1

Solution:
Applying Kirchhoff's law in closed loop,
-VR-VL=0
⇒ VR VL =- 1

Topic: Physics, Alternating Current, JEE Main, JEE Main 2014 (06 Apr)
In the circuit shown here, the voltage across $\mathrm{E}$ and $\mathrm{C}$ are respectively $300 \mathrm{~V}$ and $400 \mathrm{~V}$. The voltage $\mathrm{E}$ of the ac source is :

Options:
A
$400$ Volt

B
$500$ Volt

C
$100$ Volt

D
$700$ Volt

The correct answer is:
$100$ Volt

Solution:
Voltage $E$ of the ac source
$\begin{aligned}
& E=V_C-V_L \\
& =400 \mathrm{~V}-300 \mathrm{~V}=100 \mathrm{~V}
\end{aligned}$

Topic: Physics, Alternating Current, JEE Main, JEE Main 2013 (09 Apr Online)
In the circuit shown, if the 10 Î© resistance is replaced by 20 Î©, then the amount of current drawn from the battery will be

Options:
A 10Â A
B 4 A
C 8Â A
D 2Â A

The correct answer is:
4 A

Solution:
âµ38=924
â´ The Wheatstone bridge is balanced.
â´ On changing the resistance of the bridge, the current drawn will remain the same.

Topic: Physics, Current Electricity, NEET
In the circuit shown in diagram, switch S is closed at time t=0. The charge which passes through the battery in one time constant is

Options:
A eR2EL
B ELR
C ELeR2
D eLER

The correct answer is:
ELeR2

Solution:
The current at time t is given by
i=i01-e-t/τ
Here, i0=ER and τ=LR
∴ q=∫0τ idt=∫0τ i01-e-t/τdt
=i0τe=ERLRe
=ELeR2

Topic: Physics, Alternating Current, JEE Main
In the circuit shown in figure, the base current IB is 10 μA and the collector current is 5.2 mA. The voltage (VBE) across the base and emitter is

Options:
A 0.1 V
B 0.5 V
C 0.25 V
D 0.7 V

The correct answer is:
0.5 V

Solution:
VBE=VCC-IBRB

VBE=5.5-10Ã10-6Ã5Ã105=0.5 V

Topic: Physics, Semiconductors, JEE Main
In the circuit shown in the figure. A, R3 is a variable resistance

As the value R3 is changed, current I though the cell varies as shown. Obviously, the variation is asymptotic, i.e. I â 6 A as R3ââ . Resistances R1 and R2 are, respectively

Options:
A 4Â Î©,Â 2Â Î©
B 2Â Î©,Â 4Â Î©
C 2 Î©, 2 Î©
D 1Â Î©,Â 4Â Î©

The correct answer is:
2 Î©, 2 Î©

Solution:
For R3=0,I=9 A (from the graph).In this situation, the circuit can be drawn as shown in figure A .
[current through R2 will be zero because we have a short circuit R3=0 across R2]
Here I=36R1+R2=9 A
or R1+R2=4 Ω ...(1)
current will not flow through R 3 , it will be open circuit.
In this situation, the circuit can be drawn as shown figure B .
Figure (B)
Now I=36R1+2R2 =6 A (given)
∴ R1+2R2=6 Ω ...(2)
from Eqs. (1) and (2)
R1=2 Ω, R2=2 Ω

Topic: Physics, Current Electricity, JEE Main
In the circuit shown in the figure (A), R3 is a variable resistance. As the value R3 is changed, the current I through the cell varies as shown in the figure (B). As R3ââ, the current I â 6 A. What is the value of the sum of the resistances R1 and R2 in ohms?

Options:

The correct answer is:
4

Solution:
For R3=0,I=9 A (from the graph).In this situation, the circuit can be drawn as shown in figure A .
[current through R2 will be zero because we have a short circuit R3=0 across R2]
Here I=36R1+R2=9 A
R1+R2=4 Î©

Topic: Physics, Current Electricity, JEE Main
In the circuit shown in the figure, E=15 V, R1=1 Î©, R2=1 Î©, R3=2 Î© and L=1.5 H. The currents flowing through R1, R2 and R3 are i1, i2 and i3 , respectively.

Immediately after connecting switch Sw,

Options:
A i3=0 A and di3dt=0 A s-1
B i3=0 A and di3dtâ 0 A s-1
C i3=0 A , and rate at which magnetic energy stored is not zero
D None of these

The correct answer is:
i3=0 A and di3dtâ 0 A s-1

Solution:
At t = 0 , circuit can be considered as follows.
After a long time, circuit can be considered as follows.
The current i3 is zero at the instant of closing the switch but with time it grows and achieves a non-zero value, hence di3dt≠0

Topic: Physics, Alternating Current, JEE Main
In the circuit shown in the figure, if potential at point $A$ is taken to be zero, the potential at point $B$ is

Options:
A $-1 \mathrm{~V}$
B $+2 \mathrm{~V}$
C $-2 \mathrm{~V}$
D $+1 \mathrm{~V}$

The correct answer is:
$+1 \mathrm{~V}$

Solution:
By KVL along path $A C D B$
$$
\begin{gathered}
V_A+1+(1)(2)-2=V_B \\
0+1=V_B \\
V_B=1 \mathrm{~V}
\end{gathered}
$$

Topic: Physics, Current Electricity, NEET, NEET 2011 (Mains)
In the circuit shown in the figure, if potential at point A is taken to be zero , the potential at point B is ?

Options:
A 3 V
B 1 V
C 4 V
D 2 V

The correct answer is:
3 V

Solution:
Current From D to C = 3 A
â´ VD-VC=2Ã3=6 V
VA=0 â´ VC=-1 V
â´ VD+1=6
â VD=5 V
VD-VB=2
â´ 5-VB=2
â VB=3 V

Topic: Physics, Current Electricity, JEE Main
In the circuit shown in the figure, the current flowing in $2 \Omega$ resistance

Options:
A $1.4 \mathrm{~A}$
B $1.2 \mathrm{~A}$
C $0.4 \mathrm{~A}$
D $1.0 \mathrm{~A}$

The correct answer is:
$1.0 \mathrm{~A}$

Solution:
Current $\quad$ through $\quad 2 \quad \Omega$
$=1.4\left\{\frac{(25+5)}{(10+2)+(25+5)}\right\}=1 A$

Topic: Physics, Current Electricity, JEE Main
In the circuit shown in the figure, the ratio of the quality factor and the band width is _____ s.

Options:

The correct answer is:
10

Solution:
For an LCR circuit,

Bandwidth =RL

=103 Hz

For an LCR circuit quality factor =LRC

=310×27×10-6=1003

Therefore, required ratio

=10.

Topic: Physics, Alternating Current, JEE Main
In the circuit shown in the figure, the ratio of the quality factor and the band width is _____ s.

Options:

The correct answer is:
10

Solution:
For an LCR circuit,

Bandwidth =RL

=103 Hz

For an LCR circuit quality factor =LRC

=310×27×10-6=1003

Therefore, required ratio

=10.

Topic: Physics, Alternating Current, JEE Main, JEE Main 2023 (24 Jan Shift 1)
In the circuit shown in the figure, the reading of the voltmeter is V1 when only S1 is closed, V2 when only S2 is closed and V3 when both S1 and S2 are closed. From this we can conclude that

Options:
A V3>V2>V1
B V2>V1>V3
C V3>V1>V2
D V1>V2>V3

The correct answer is:
V2>V1>V3

Solution:
In series V âR
When S1 is closed and S2 is open
V1=3E4
When S1 is open and S2 is closed
V2=6E7
When S1 and S2 both are closed
V3=2E3

Topic: Physics, Current Electricity, JEE Main
In the circuit shown in the figure, the switch was kept in position-1 for a very long time and then at t=0 it is shifted to position-2. The current in the circuit immediately after that is i=ab A, then the value of a+b is

Options:

The correct answer is:
7

Solution:
The total flux, before and after the position change of the switch, remains the same

L1i1=L2i2

L1=5 H, i1=4 A and L2=15 H

5Ã4=15Ãi

âi=43=ab

âa+b=7

Topic: Physics, Alternating Current, JEE Main
In the circuit shown in the figure, the total charge is 750 μC and the voltage across capacitor C2 is 20 V. Then the charge on capacitor C2 is :

Options:
A 450 μC
B 590 μC
C 160 μC
D 650 μC

The correct answer is:
590 μC

Solution:
Applying Kirchoff's law we get,

q2+q3=750

q2+160=750

q2=590 μC

Topic: Physics, Capacitance, JEE Main, JEE Main 2020 (03 Sep Shift 1)
In the circuit shown in the figure, the voltmeter reading is:

Options:
A \( 2 \mathrm{~V} \)
B \( 2.5 \mathrm{~V} \)
C \( 3 \mathrm{~V} \)
D \( 5 \mathrm{~V} \)

The correct answer is:
\( 2 \mathrm{~V} \)

Solution:
Given,

emf of the cell, V=5 V

The voltmeter200 Ω and 100 Ω resistor are in parallel and their equivalent will be in series with the another 100 Ω resistor.

The equivalent resistance of the given network is

Req=100+100×200100+200

Req=100+2003=5003 Ω

Current through the circuit is,

i=VReq=55003=3100 A

The voltage reading of the voltage across 100 Ω resistor connected with voltmeter, which is same as the potential difference of equivalent resistance of 2003 Ω.

V'=3100×2003=2 V.

Topic: Physics, Current Electricity, JEE Main
In the circuit shown, $\mathrm{n}$ identical resistors $\mathrm{R}$ are connected in parallel $(\mathrm{n}>1$ ) and the combination is connected in series to another resistor $\mathrm{R}_{0}$. In the adjoining circuit $\mathrm{n}$ resistors of resistance $\mathrm{R}$ are all connected in series along with $\mathrm{R}_{0}-$
The batteries in both circuits are identical and net power dissipated in the n resistors in both circuits is same. The ratio $\mathrm{R}_{0} / \mathrm{R}$ is

Options:
A 1
B $\mathrm{n}$
C $\mathrm{n}^{2}$
D $1 / \mathrm{n}$

The correct answer is:
1

Solution:
$\begin{array}{l}
\mathrm{i}_{1}=\frac{\mathrm{nE}}{\mathrm{n} R_{0}+\mathrm{R}}, \mathrm{i}_{2}=\frac{\mathrm{E}}{\left(\mathrm{R}_{0}+\mathrm{n} \mathrm{R}\right)} \\
\mathrm{P}_{1}=\frac{\mathrm{nE}^{2} \mathrm{R}}{\left(\mathrm{nR}_{0}+\mathrm{R}\right)^{2}}, \mathrm{P}_{2}=\frac{\mathrm{nE}^{2} \mathrm{R}}{\left(\mathrm{R}_{0}+\mathrm{nR}\right)^{2}} \\
\because \mathrm{P}_{1}=\mathrm{P}_{2} \\
\text { Hence } \mathrm{R}_{0} / \mathrm{R}=1
\end{array}$

Topic: Physics, Current Electricity, JEE Main
In the circuit shown switch S is connected to position 2 for a long time and then joined to the position 1. The total heat produced in resistance R1 is :

Options:
A LE 2 2 R 2 2
B LE 2 2 R 1 2
C LE 2 2 R 1 R 2
D LE 2 R 1 + R 2 2 2 R 1 2 R 2 2

The correct answer is:
LE 2 2 R 2 2

Solution:
When the key is at position (B) for a long time; the energy stored in the inductor is :
U B = 1 2 Li 0 2 = 1 2 · L · E R 2 2 = LE 2 2 R 2 2

This whole energy will be dissipated in the form of heat when the inductor is connected to R1 and no source is connected.

Topic: Physics, Electromagnetic Induction, JEE Main
In the circuit shown, the current flowing from the battery is

Options:
A 1150 A
B 10Â A
C 53 A
D Infinite current

The correct answer is:
10 A

Solution:
By observing the reduced/equivalent circuit diagram, we can say that,
i=VR=202=10 A

Topic: Physics, Current Electricity, JEE Main
In the circuit shown, the current in the 1 Ω resistor is:

Options:
A 0.13 A, from P to Q
B 1.3 A, from P to Q
C 1.3 A, from Q to P
D 0.13 A, from Q to P

The correct answer is:
0.13 A, from Q to P

Solution:
R=3 Ω+2 Ω. Let potential at P=0, ∴VA=6 V, VD=−9 V, andVQ=x. Apply K.C.L. at x, x−6 3 + x−0 1 + x+9 5 =0 . Solving, x=323 V. So, current in 1 Ω⇒i=323−01 from Q to P⇒i=0.13 A from Q to P

Topic: Physics, Current Electricity, JEE Main, JEE Main 2015 (04 Apr)
In the circuit shown, the current through the $4 \Omega$ resistor is $1 \mathrm{amp}$ when the points $P$ and $\mathrm{M}$ are connected to a d.c. voltage source. The potential difference between the points $M$ and $N$ is

Options:
A 3.2 volt
B 1.5 volt
C 1.0 volt
D 0.5 volt

The correct answer is:
3.2 volt

Solution:
Potential difference between $P$ and $M=1 \times 4=4$ volt
Potential drop between points $M$ and $N=\frac{4 \times 1}{1+0.25}=3.2 \mathrm{~V}$

Topic: Physics, Current Electricity, JEE Main
In the circuit shown, the current through the $4 \Omega$ resistor is $1 \mathrm{amp}$ when the points $P$ and $\mathrm{M}$ are connected to a d.c. voltage source. The potential difference between the points $M$ and $N$ is

Options:
A 3.2 volt
B 1.5 volt
C 1.0 volt
D 0.5 volt

The correct answer is:
3.2 volt

Solution:
Potential difference between $P$ and $M=1 \times 4=4$ volt
Potential drop between points $M$ and $N=\frac{4 \times 1}{1+0.25}=3.2 \mathrm{~V}$

Topic: Physics, Current Electricity, NEET, NEET 2008 (Mains)
In the circuit shown, the heat produced in 5 Î© resistance due to current through it is 50 J s-1. Then, the heat generated per second in the 2 Î© resistance is

Options:
A 5 J s-1
B 4 J s-1
C 9 J s-1
D 10 J s-1

The correct answer is:
5 J s-1

Solution:
Power is given as, P=50 J s-1
P=Vi ⇒P=i2R

⇒ i22=PR=505=10 A2
V=i2R5 Ω
= 10×5= 250 V
and 2 Ω and 8 Ω are in series.
So, the required resistance =2 Ω+8 Ω=10 Ω
i1=VR10 Ω
i1=25010 A
The heat generated per second in 2 Ω is given by,
P=Vi1
=i12×R
=250102×2
=250100×2=2510×2=255=5 J s-1

Topic: Physics, Current Electricity, JEE Main
In the circuit shown, the potential difference between $\mathrm{A}$ and $B$ is

Options:
A $1 \mathrm{~V}$
B $2 \mathrm{~V}$
C $3 V$
D $6 \mathrm{~V}$

The correct answer is:
$2 \mathrm{~V}$

Solution:
Given, $E_{1}=1 \mathrm{~V}, \mathrm{E}_{2}=2 \mathrm{~V}, \mathrm{E}_{3}=3 \mathrm{~V}, \mathrm{r}_{1}=1 \Omega$

$\mathrm{r}_{2}=1 \Omega$ and $\mathrm{r}_{3}=1 \Omega$

$\mathrm{V}_{\mathrm{AB}}=\mathrm{V}_{\mathrm{CD}}=\frac{\frac{\mathrm{E}_{1}}{\mathrm{r}_{1}}+\frac{\mathrm{E}_{2}}{\mathrm{r}_{2}}+\frac{\mathrm{E}_{3}}{\mathrm{r}_{3}}}{\frac{1}{\mathrm{r}_{1}}+\frac{1}{\mathrm{r}_{2}}+\frac{1}{\mathrm{r}_{3}}}=\frac{1}{\frac{1}{1}+\frac{1}{1}+\frac{1}{1}+\frac{3}{1}}=\frac{6}{3}=2 \mathrm{~V}$

Topic: Physics, Current Electricity, JEE Main, JEE Main 2019 (11 Jan Shift 2)
In the circuit shown, the resistance of each resistor is R=100 Î©. Find the resistance in Î© to the nearest integer, which is to be connected between points C and D so that the resistance of entire circuit between A and B does not depend on the number of elementary cells.

Options:

The correct answer is:
73

Solution:
Let resistance added across CD is x, then the equivalent resistance between A and B will be independent of the number of elementary cells if the combination of x with the resistors of the elementary cells is x itself. So RAB=x
∴ x=2R+x×R3R+x
⇒ x=3-1×R
=1.73-1×100
=73 Ω

Topic: Physics, Current Electricity, JEE Main
In the circuit shown, the resistance r is a variable resistance. If for r=fR , the heat generation in r is maximum then the value of f is

Options:
A 12
B 1
C 14
D 34

The correct answer is:
12

Solution:
Power in “r”
P= I 2 2 r= [ I× R R+r ] 2 r= I 2 R 2  r ( R+r ) 2
P= [ E R+ R×fR R+fR ] 2 R 2 ( fR ) ( R+fR ) 2 = E 2 R 2 fR R 2 [ 1+ f ( f+1 ) ] 2 R 2 ( 1+f ) 2
P= E 2 f R ( 1+2f ) 2
For P( max )  dP df =0
E 2 R =[ ( 1+2f ) 2 −f{ 2( 1+2f )( 0+2 ) } ( 1+2f ) 2 ]=0
( 1+2f ) 2 =4f( 1+2f )
1+2f=4f
f= 1 2

Topic: Physics, Current Electricity, JEE Main, JEE Main 2016 (09 Apr Online)
In the circuit shown,
the switch $S_{1}$ is closed at time $t=0$ and the switch $S_{2}$ is kept open. At some later time $\left(\mathrm{t}_{0}\right)$, the switch $\mathrm{S}_{1}$ is opened and $\mathrm{S}_{2}$ is closed. the behaviour of the current I as a function of time ' $\mathrm{t}$ ' is given by:

Options:
A
B
C
D

The correct answer is:

Solution:
The current will grow for the time $\mathrm{t}=0$ to $\mathrm{t}=\mathrm{t}_{0}$ and after that decay of current takes place.
Growth and decay of current is of exponential nature
$\mathrm{i}=\mathrm{i}_0\left(1-\mathrm{e}^{-\mathrm{t} / \mathrm{\tau}}\right) \rightarrow \text { during growth }$
$\mathrm{i}=\mathrm{i}_{\max } \mathrm{e}^{-\mathrm{t} / \mathrm{\tau}} \rightarrow \text { during decay }$

Topic: Physics, Alternating Current, JEE Main, JEE Main 2019 (11 Jan Shift 1)
In the circuit shown, the value of $\beta$ of the transistor is 48 . If the base current supplied is $200 \mu \mathrm{A}$, what is the voltage at the terminal $\mathrm{Y}$?

Options:
A $0.2 \mathrm{~V}$
B $0.5 \mathrm{~V}$
C $4 \mathrm{~V}$
D $4.8 \mathrm{~V}$

The correct answer is:
$0.2 \mathrm{~V}$

Solution:
Hint:
$\beta=48$
$l_{B}=200 \mu \mathrm{A}$
$I_{C}=\beta I_{B}=48 \times 200 \times 10^{-6}$
$V_{\mathrm{CC}}=I_{C} R_{C}+V_{C E} \therefore V_{C E}=5-\left(96 \times 10^{-4}\right) \times 500=5-4.8=0.2$ volt

Topic: Physics, Semiconductors, JEE Main
In the circuit shown, What is the current (in A) in the 1 Î© resistor?

Options:

The correct answer is:
0.13

Solution:
R=3 Î©+2 Î©
Let Potentinal at P = 0
â´VA=6 V,âVD=â9 V
and VQ=x
Apply KCL at x
xâ6 3 + xâ0 1 + x+9 5 =0
Solving, x=323 V

So, current in 1 Î© resistance is
i= 3 23 â0 1 from Q to P
i = 0.13 A from Q to P

Topic: Physics, Current Electricity, JEE Main
In the circuit shows in the figure, the input voltage Vi is 20 V, VBE=0 and VCE=0. The values of IB,IC and Î² are given by

Options:
A IB=20 Î¼A, IC=5 mA, Î²=250
B IB=25Â Î¼A,Â IC=5Â mA,Â Î²=200
C IB=40Â Î¼A,Â IC=10Â mA,Â Î²=250
D IB=40 Î¼A, IC=5 mA, Î²=125

The correct answer is:
IB=40 Î¼A, IC=5 mA, Î²=125

Solution:
Vi=IBRB+VBE (By Kirchoff's Voltage Law)
20=IBÃ500Ã103+0
IB=20500Ã103=40 Î¼A
VCC=ICRC+VCE (By Kirchoff's Voltage Law)
20=ICÃ4Ã103+0
IC=5Ã10-3=5 mA
Î²=ICIB=5Ã10-340Ã10-6=125

Topic: Physics, Semiconductors, JEE Main
In the circuit, the galvanometer $G$ shows zero deflection. If the batteries $A$ and $B$ have negligible internal resistance, the value of the resistor $\mathrm{R}$ will be

Options:
A
$200 \Omega$

B
$100 \Omega$

C
$500 \Omega$

D
$1000 \Omega$

The correct answer is:
$100 \Omega$

Solution:
$\frac{12 \mathrm{R}}{500+\mathrm{R}}=2$

Topic: Physics, Current Electricity, JEE Main, JEE Main 2005