Physics

Current Electricity and Circuits

377 Questions

Current electricity and circuits questions cover resistors, EMF, internal resistance, and power calculations in series and parallel configurations. Solving these builds a strong understanding of electrical principles and circuit analysis. These physics problems are highly relevant for technical and science aptitude tests.

Resistor combinationsPower dissipationEMF and internal resistanceAC circuit analysisOperational amplifiers

Current Electricity and Circuits Questions

Multiple choice physics electric current through conductors combination of resistors combination of cells combinations of components

Two cells, having the same e.m.f are connected in series through an external resistance R. Cells have internal resistance $r _1$ and $r _2$ ( $r _1$ > $r _2$) respectively. When the circuit is closed, the potential difference across the first cell is zero. the value of R is

  1. $\dfrac{r _1 + r _2}{2}$
  2. $\dfrac{r _1 - r _2}{2}$
  3. ${r _1 + r _2}$
  4. ${r _1 - r _2}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
Two cells of emf E are connected in series through external resistance R.
Thus, total resistance is R + r1 + r2
 
Thus, current across the circuit is total voltage divided by total resistance.
I=E+E/R+r1+r2
I=2E/R+r1+r2
 
Now, when circuit is closed, the voltage across first cell is zero. Thus, we have
Ir1=E
E/r1=2E/R+r1+r2
R=r1-r2
This is the value of R in terms of r1 and r2.
Multiple choice physics electric current through conductors combination of resistors combination of cells combinations of components

A battery of 20 cells (each having e.m.f. 1.8 volt and internal resistance 0.1 ohm) is charged by 220 volts and the charging current is 15A. The resistance to be put in the circuit is

  1. 10.27 ohms

  2. 12.27 ohms

  3. 8.62 ohms

  4. 16.24 ohms
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The total emf of the 20 cells connected in series are $1.8\times 20=36 V$. 

Hence, the total voltage in the circuit is $220V - 36V  =  184 V.$
From the Ohm's law, $V=IR$. Substituting the values of Voltage and current in the equation

 $R=\dfrac { V }{ I } ,\quad we\quad get\quad R=\dfrac { 184 }{ 15 } =12.27\quad ohms$.

The internal resistance of the 20 cells is given as  $20\times 0.1ohms=2 ohms$.
So, the total resistance R in the circuit is $ =12.27-2 ohms = 10.27 ohms.$
Hence, the resistance in the circuit is 10.27 ohms.

Multiple choice physics electric circuits combination of resistors combination of cells combinations of components

A simple electric motor has an armature resistance of $1\Omega$ and runs from a d.c. source of $12$V. It draws a current of $2$A when unloaded. When a certain load is connected to it, its speed reduces by $10\%$ of its initial value. The current drawn by the loaded motor is?

  1. $3$A
  2. $6$A
  3. $2$A
  4. $1$A
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The back emf$= V-IR$ 
$= 12-2 = 10V$
$10%$ of this emf is $1V$
Hence the current is $IR = 1\times1= 1A$
Hence the current drawn by the loaded motor is $1+2=3A$

Multiple choice physics electric circuits combination of resistors combination of cells combinations of components

When two identical cell are connected either in series or in parallel across a 4 ohm resistor, they send the same current through it. The internal resistance of the cell in ohm is:

  1. 1.2

  2. 2

  3. 4

  4. 4.8
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

For identical cells, the current in series is E/(R + nr) and in parallel is E/(R + r/n). Setting these equal for R=4 gives r=4.

Multiple choice physics electric circuits combination of resistors combination of cells combinations of components

A series battery of six lead accumulators, each of emf 2.0V and internal resistance 0.50$\Omega $ is charged by a 100 V dc supply. The series resistance should be used in the charging circuit in order to limit the current to 8.0A is

  1. 4$\Omega $
  2. 6$\Omega $
  3. 8$\Omega $
  4. 10$\Omega $
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$ \Rightarrow  (100-12) = I[3+R]$

$ \dfrac{88}{3+R} = 8 $

$ \Rightarrow  R = 8\Omega$

Multiple choice physics electric circuits combination of resistors combination of cells combinations of components

A thin uniform wire $50\ cm$ long and of $1\ ohm$ resistance is connected to the terminals of an accumulator of $emf\ 2.2\ volt$ and the internal resistance $0.1\ ohm$. If the terminal of another cell can be connected to two point $26\ cm$ apart on the wire without altering the current in the wire, the emf of the cell is

  1. $1.12\ V$
  2. $1.04\ V$
  3. $1.18\ V$
  4. $1.22\ V$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Using the potentiometer principle, the potential drop across 26 cm of the wire is proportional to the EMF of the cell. Calculation: (2.2 * 26/50) / (1 + 0.1) * (1) = 1.22V.

Multiple choice physics electric circuits combination of resistors combination of cells combinations of components

Two cells of same emf are connected in series. Their internal resistances are $r _1$ and $r _2$ respectively and $r _1 > r _2$. When this combination is connected to an external resistance R then the potential difference between the terminals of first cell becomes zero. In this condition the value of R will bw

  1. $\frac {r _1-r _2}{2}$
  2. $\frac {r _1+r _2}{2}$
  3. $r _1-r _2$
  4. $r _1+r _2$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Let emf of each cell is $E$.
As they are connected in series so current in circuit is $I=\dfrac{E+E}{r _1+r _2+R}=\dfrac{2E}{r _1+r _2+R}$ 

Potential across terminal of first cell is $V _1=E-Ir _1=E-\dfrac{2Er _1}{r _1+r _2+R}$ 

As $V _1=0 \Rightarrow E-\dfrac{2Er _1}{r _1+r _2+R}=0$ 

$r _1+r _2+R-2r _1=0$ 

$R=r _1-r _2$

Multiple choice physics electric circuits combination of resistors combination of cells combinations of components

$24$ cells, each having the same e.m.f. and $2$ ohm internal resistance, are used to draw maximum current through an external resistance of $3$ ohm. The cells should be connected :

  1. In series

  2. in parallel
  3. In $4$ rows, each row having $6$ cells
  4. In $6$ rows, each row having $4$ cells
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

(A) Option in series 

Net EMF = $n\varepsilon$
= 24$\epsilon$
Net resistance =nr
= 24$\times$2
=48$\Omega$
Current = $\dfrac { 24\varepsilon  }{ nr+R } $
$ =\dfrac { 24\varepsilon  }{ nr+R } $
$=\dfrac { 24\varepsilon  }{ 48+3 } $
 $=\dfrac { 24\varepsilon  }{ 51 } =0.47\varepsilon $
(B) Net EMF = $\varepsilon$ = (In parallel EMF is same )
Net resistance $=\dfrac { r }{ n } =\dfrac { 1\Omega  }{ 12 } $
Current $=\dfrac { \varepsilon  }{ \dfrac { 1 }{ 12 } +3 } $
$=\dfrac { 12\varepsilon  }{ 37 } =0.32\varepsilon $
(C) Option
net EMF in each row$ = n\varepsilon$
$= 6\varepsilon$
net resistance in each row $=n\varepsilon$
$= 12\Omega$
net EMF will be $6\varepsilon$ ( this will be in parallel)
Net resistance $\dfrac{r}{n}$
$=\dfrac{12}{4}=3\Omega$
Net current $=\dfrac{6\varepsilon}{3+3}$
$= 1\varepsilon A$
(D) Option 
Net EMF in each row $n\varepsilon$
=$4\varepsilon$
Net resistance in each row = nr
= 4$\times$2
= 8$\Omega$
Net EMF will be $4\varepsilon$ ( this will be in parallel)
Net resistance $\dfrac{r}{n}=\dfrac{8}{6}=\dfrac{4}{3}\Omega$
Current $ \dfrac { 4\varepsilon  }{ \dfrac { 4 }{ 3 } +3 } =\dfrac { 12\varepsilon  }{ 13 } =0.92\varepsilon $
Hence current will be maximum in (C) option

Multiple choice physics electric circuits combination of resistors combination of cells combinations of components

n identical cells are joined in series with its two cells A and B in the loop with reversed polarities.
EMF of each shell is E and internal resistance r. Potential difference across cell A or B is:

  1. $\dfrac{2E}{n}$
  2. $2E|1-\dfrac{1}{n}|$
  3. $\dfrac{4E}{n}$
  4. $2E|1-\dfrac{2}{n}|$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Number of cells $=n$

EMF of each cell $=E$
Internal resistance of each cell, $= r$

All the cells are in series.
Thus,
Total internal resistance, $R=nr$
Total EMF, $e=nE$

As the two cells are connected in reverse polarity, these two cells will cancel-out the contribution of other two cells in the loop.

Then, the net EMF across the circuit will be
$E _{eq}=nE-4E$

The current $I$ in the circuit $=\dfrac{nE-4E}{nr}$

Voltage across the opposite connected batteries, $V=E-Ir$
$V=E-\dfrac{E(n-4)}{nr}\times r$
$V=\dfrac{4E}{n}$

Therefore, the voltage across the battery A or B will be $\dfrac{4E}{n}$.


Multiple choice physics electric circuits combination of resistors combination of cells combinations of components

Two similar cells, whether joined in series or in parallel, have the same current through an external resistance of $2\Omega$. The internal resistance of each cell is

  1. $1\Omega$
  2. $2\Omega$
  3. $0.5\Omega$
  4. $1.5\Omega$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

In series, current, $i _{1} = \dfrac {2E}{2 + 2r}$
In parallel, current, $i _{2} = \dfrac {E}{2 + \dfrac {r}{2}} = \dfrac {2E}{4 + r}$
According to the question
Since, $i _{1} - i _{2} \Rightarrow \dfrac {2E}{4 + r} = \dfrac {E}{2 + 2r}$
$\Rightarrow r = 2\Omega$.

Multiple choice physics electric circuits combination of resistors combination of cells combinations of components

36 identical cell each having emf 1.5 volt and internal resistance $ 0.5 \Omega$ are connected in series with an external resistance of $12 \Omega$ .If 6 cells are wrongly connected then current through the circuit will be 

  1. 1.2 A

  2. 1 A

  3. 2 A

  4. 4 A
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Total resistance in the circuit is, $R _T=36\times0.5+12=30\Omega$

Net EMF in the circuit is, $E _T=(30-6)\times1.5=36V$

So, current in the circuit is, $I=\dfrac{E _T}{R _T}=\dfrac{36}{30}=1.2A$

Multiple choice physics electric circuits combination of resistors combination of cells combinations of components

When n number identical cell of emf E and internal resistance is connected in series, the net internal resistance of the system will be  

  1. $\dfrac{nEr}{1+E}$
  2. $n^2r$
  3. $nr$
  4. $n/r$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

It is a common fact that when current flows the circuit, connected with a magnetic compass, the magnetic needle in the compass will show deflection due to the magnetic effect of the current. Magnetic lines of forces are created around the coil of wire and this caused the needle to deflect according to the direction of the current.
In this case, when the number of cells in the circuit is increased, the current through the circuit is also increased. This is because, the increase in cells indicates more current in the circuit. 
Hence, as the current in the circuit is increased, the deflection in the magnetic compass also increase further.

Multiple choice physics electric circuits combination of resistors combination of cells combinations of components


A cell of constant emf first connect to a resistance $R _1$ and then to connected to the resistance $R _2.$ If power delivered in both cases in the same then internal resistance of the cell

  1. $\dfrac { R _ { 1 } - R _ { 2 } } { 2 }$
  2. $ { R _ { 1 } + R _ { 2 } } $
  3. $\sqrt { R _ { 1 } R _ { 2 } }$
  4. $\sqrt { R _ { 1 } + R _ { 2 } / 2 }$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The cell has constant emf$,$

Resistance is $R _1$ and $R _2$

$P = V _2 / R _1 + r$

$P = V _2 / R _2 - r$

$( R _1- R _2 ) / 2$

The internal resistance of the cell is $( R _1 - R _2 ) / 2$

Hence,

option $(A)$ is correct answer.