Tag: physics

Questions Related to physics

When cells are connected in series greater emf is obtained in the circuit.

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

  1. True

  2. False

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Correct Option: A
Explanation:

When two or more cells are connected in series in a circuit, the total potential difference is the sum of their potential differences. Hence, the cells are connected in series in order to produce a larger total emf.

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

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

  1. $\dfrac{r _1 + r _2}{2}$

  2. $\dfrac{r _1 - r _2}{2}$

  3. ${r _1 + r _2}$

  4. ${r _1 - r _2}$

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Correct Option: D
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.

Which of the following statement(s) is/are correct?

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

  1. If n identical cells are connected in series and then the battery thus formed is short circuited by a conduction wire, current through the wire will be independent of n.

  2. If n identical cells are connected in parallel and then the battery thus formed is short circuited by a conducting wire, current through the wire will be directly proportional to n.

  3. If n identical cells are connected in parallel and then the battery thus formed is short circuited by a wire having a constant resistance, current through the wire will be directly increase as n increases.

  4. None of the above.

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Correct Option: A,B
Explanation:

In series, $I=\frac{nE}{nr}=\frac{E}{r}$ (i.e., I is 
independent of n)

In parallel, $I=\frac{E}{r/n}=\frac{nE}{r}, i.e,.\frac{nE}{r}, 
i.e.I\propto n$

After shorting by a wire having a constant resistance 
$I=E\left (R+\frac{r}{n}  \right )$

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

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

  1. 10.27 ohms

  2. 12.27 ohms

  3. 8.62 ohms

  4. 16.24 ohms

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Correct Option: A
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.

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?

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

  1. $3$A

  2. $6$A

  3. $2$A

  4. $1$A

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Correct Option: A
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 $1+2=3A$

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:

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

  1. 1.2

  2. 2

  3. 4

  4. 4.8

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Correct Option: C

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

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

  1. 4$\Omega $

  2. 6$\Omega $

  3. 8$\Omega $

  4. 10$\Omega $

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Correct Option: C
Explanation:

$ \Rightarrow  (100-12) = I[3+R]$

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

$ \Rightarrow  R = 8\Omega$

$n$ cells of each of EMF $E$ and internal resistance $r$ send the same current $R$ whether the cells are connected in series or parallel, then :

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

  1. $R = nr$

  2. $R = r$

  3. $r = nR$

  4. $ R = \sqrt {n}  \dfrac{E}{r}$

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Correct Option: D

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

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

  1. $1.12\ V$

  2. $1.04\ V$

  3. $1.18\ V$

  4. $1.22\ V$

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Correct Option: D

Two non-ideal batteries are connected in series. Consider the following statements:
(A) The equivalent emf is larger than either of the two emfs.
(B) The equivalent internal resistance is smaller than either of the two internal resistances

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

  1. Each of A and B is correct

  2. A is correct but B is wrong

  3. B is correct but A is wrong

  4. Each of A and B is wrong

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Correct Option: B
Explanation:

When two non-ideal batteries are connected in series, their equivalent emf is
$E _{eq} = E _1 + E _2$
Hence, the equivalent emf is larger than either of the two emfs.
Also, their equivalent internal resistance is
$r _{eq} = r _1 + r _2$
Hence, the equivalent resistance is larger than either of the two internal resistances.
Therefore, statement A is correct but B is wrong.