# Important Short Questions With Answers – DC Circuit (Direct Current)| Class 12 Physics

### Q. A wire is stretched to double its length. What happens to its resistance and resistivity?

Ans: Since the resistivity of the wire depends upon the nature of the material of the conductor, therefore the resistivity remains same for the same conductor. Let us consider a wire of length l, resistance R, resistivity and area of cross section A. Then resistance

R = $\frac{\rho l}{A}$

Or, R = $\frac{\rho {{l}^{2}}}{Al}$

Or, R = $\frac{\rho {{l}^{2}}}{V}$ …………. (i), where Al = V is the volume of the conductor

If the wire is stretched to double its length, then the new length is

l’ = 2l and the new resistance R’ can be written as

R = $\frac{\rho l’ {^{2}}}{V’}$ = $\frac{\rho {{(2l)}^{2}}}{V}$ ……………(ii), where V = V’ because volume remains constant.

Also $\rho$ = $\rho’$, because the material of the conductor is same

Dividing (ii) by (i)

R’ = R 22 = 4 R

Thus, when a wire is stretched to double its length, its resistance becomes 4 times of the original resistance.

### Q. You are given n wires each of resistance ‘R’ . What is the ratio of maximum to minimum resistance that can be obtained from these wires?

Ans: Since in series combination effective resistance increases, therefore, when n wires each of resistance R connected in series the maximum resistance is obtained is given by

Rs = R + R + …. ….upto n terms

Or, Rs = n R ………………(i)

In parallel combination effective resistance decreases, so minimum resistance is given by

$\frac{1}{{{R}_{p}}}$ = $\frac{1}{R}$ + $\frac{1}{R}$ + …. upto n terms = $\frac{n}{R}$

Or, $\frac{1}{{{R}_{p}}}$ = $\frac{n}{R}$

Or, Rp = $\frac{R}{n}$…….(ii)

Where Rs and Rp be the effective resistance in series and parallel combination.

Dividing Eq. (i) by Eq. (ii), we get

$\frac{{{R}_{s}}}{{{R}_{p}}}$ = $\frac{nR}{R/n}$ = n2

Thus required ratio, Rmax : Rmin= n2:1

### Q. You are given 2 wires each of resistance ‘R’ . What is the ratio of maximum to minimum resistance that can be obtained from these wires?

Ans: Since in series combination effective resistance increases, therefore, when 2 wires each of resistance R connected in series the maximum resistance is obtained is given by

Rs = R + R

Or Rs = 2R ……………… (i)

In parallel combination effective resistance decreases, so minimum resistance is given by

$\frac{1}{{{R}_{p}}}$ = $\frac{1}{R}$ + $\frac{1}{R}$

Or, $\frac{1}{{{R}_{p}}}$ = $\frac{2}{R}$

Or, Rp = $\frac{R}{2}$ ……………(ii)

Where Rs and Rp be the effective resistance in series and parallel combination.

Dividing Eqn. (i) by Eq. (ii), we get

$\frac{{{R}_{s}}}{{{R}_{p}}}$ = $\frac{2R}{R/2}$ = 22 = 4

Thus required ratio, Rmax : Rmin = 4:1

### Q. Why ammeter is always connected in series?

Ans: Ammeter is an instrument which is to measure current flowing in a wire. It is necessary to connect the ammeter in series so that the current to be measured passes through it. Ideally, the resistance of an ammeter is zero so that the equivalent resistance of the dc circuit does not change and hence doesn’t change the current in the circuit. In practice, the resistance is not zero but can be very low as compared to the other resistors.

If ammeter is connected in parallel, the equivalent resistance decreases drastically which increases the current in the dc circuit.

### Q. Why a voltmeter is always connected in parallel with the load resistance?

Ans: A voltmeter is a high resistance instrument which is used to measure the potential difference. When it is connected in parallel across any element of a circuit, it draws a very small amount of current from the main circuit and hence current in the circuit is altered insignificantly due to the connection of voltmeter. Most of the current passes through the element across which it is connected. Hence, p.d across the element is not affected.

If the voltmeter is connected in series, the resistance of the dc circuit becomes high. The current decreases considerably and voltmeter will not read the required p.d.

### Q. Explain the difference between resistance and resistivity of a wire.

Ans: Resistance is the property of a wire to oppose the flow of charge through it. Its unit is $\Omega$(ohm).

From Ohm’s law, V = IR

Or, R = $\frac{V}{I}$

Resistivity: From the relation, R = $\rho $$\frac{l}{A} If l = 1m, A = 1m2 then R = \rho Hence, the resistivity of the material of a conductor is defined as the resistance of unit cross – sectional area and unit length. Its unit is Ohm-meter (m). The value of resistivity of the wire is independent of its dimension. ### Q. What are the factors on which the resistance of a conductor depends? Ans: The resistance of a conductor depends upon (i) nature of material (i.e. resistivity,\rho ), (ii) length, (iii) area of cross-section and (iv) temperature. We know that R = \rho$$\frac{l}{A}$ where R is the resistance of a wire of length l and cross-sectional area ‘A’ and ” $\rho$is the resistivity of the material of the wire. From the above relation it can be concluded that

(i) R $\propto$ l

(ii) R$\propto$ $\frac{1}{A}$

So, Vd $\propto $$\frac{1}{{{d}^{2}}} So when diameter decreases, the drift velocity increases. ### Q. Will the drift velocity of electrons change if the diameter of a connecting wire is halved? Why? Ans: We have the relation between current and drift velocity I = Vd e n A Case:1 I = V1 e n A1 I = V1 e n \frac{\pi {{d}_{1}}^{2}}{4}……………(1) Case:2 when diameter is halved i.e. d = d/2 I = V2 e n A2 I = V2 e n \frac{\pi {{d}_{2}}^{2}}{4}……………(2) Dividing eqn (2) by (1), \frac{{{V}_{2}}{{(d/2)}^{2}}}{{{V}_{1}}{{d}^{2}}}= 1 V2 = 4 V Thus when diameter is halved the drift velocity increases to 4 times the initial drift velocity. ### Q. Two wires, of one of copper and another of iron, have the same diameter and carry the same current. In which wire the drift velocity of electrons will be more? Ans: We have the relation between current and drift velocity I = Vd e n A For same value of area of cross-section (or diameter) and current, we can write Vd \propto$$\frac{1}{n}$

Thus the drift velocity of electrons will be more in the wire in which less number of free electrons per unit volume.

Furthermore, the electronic configuration of Cu is

Cu (29): 1S2 2S2 2P6 3S2 3P6 4S1 3d10 i.e. one Cu-atom contributes 1 free electron.

and the electronic configuration of Fe is

Fe (26): 1S2 2S2 2P6 3S2 3P6 4S2 3d6 i.e. one Fe-atom contributes 2 free electrons.

Thus we can conclude that nFe >nCu and hence VCu > VFe i.e drift velocity of electrons is greater in copper than in iron.

### Q. Why is it essential that the resistance of a voltmeter be very high?

Ans:

A voltmeter is a device which is used to measure the potential difference across a resister. It is always connected in parallel with the resistor. The resistance of voltmeter is essential to be very high because when a very high value resistor is connected in parallel, the equivalent resistance of the circuit doesn’t decrease significantly and hence doesn’t increase the current significantly in the circuit.

### Q. The resistance of an ammeter must essentially be very small. Why?

Ans:

An ammeter is a device which is used to measure the current in the circuit. It is always connected in series with the resistor. The resistance of ammeter is essential to be very low because when a very low value resistor is connected in series, the equivalent resistance of the circuit doesn’t increase significantly and hence doesn’t decrease the current significantly in the circuit.

### Q. A cylindrical rod has resistance R. If we double its length and diameter, what is its resistance in terms of R’?

Ans:

The resistance of a wire is given by,

R = $\frac{\rho l}{A}$ = $\frac{\rho l}{\pi d{{{}_{1}}^{2}}/4}$ ………… (i)

When the length and diameter is doubled i.e. new length = 2l and diameter = 2d then new resistance R’ becomes R’ = $\frac{\rho \times 2l}{\pi {{(2d)}^{2}}/4}$ …………. (ii)

Dividing (ii) by (i) $\frac{R’}{R}$ = $\frac{1}{2}$

R’ = $\frac{1}{2}$ R.

Thus if we double the length and diameter of the wire, the resistance becomes half than that of initial resistance.

### Q. Resistors R1 and R2 are connected in parallel to an emf source that has negligible internal resistance. What happens to the current through R1 when R2 is removed from the circuit?

Ans:

When R1 and R2 are connected in parallel The current through R1 is I1 = $\frac{{{V}_{1}}}{{{R}_{1}}}$ since V = IR

Or, I1 = $\frac{E}{{{R}_{1}}}$since in parallel combination V1, V2, E are equal.

When R2 is removed The current through R1 is I1 = $\frac{E}{{{R}_{1}}}$ i.e. I1 = I1

The current through R1 remains same when R2 is removed dc circuit. But total current decreases.

Thus, we can conclude that in parallel combination, the current through a resistor remains same if we remove or add other resistor(s) parallel to the given resistor.

### Q. Resistors R1 and R2 are connected in series to an emf source that has negligible internal resistance. What happens to the current through R1 when a third resistor R3 is connected in parallel with R2?

Ans:

When a third resister R3 is connected in parallel with R2 then equivalent resistance of the circuit decreases and hence the current in the circuit (i.e. the current through R1) increases.

### Q. A large number of electrons are present in metals. Why is there no current in the absence of electric field across it?

Ans:

The electric current is due flow of charge in particular direction. But this does not happen in the above case. Though a large number of electrons are present in metals, in the absence of electric field, these free electrons have random motion. During such motion, they collide with positive ions of the metal again and again. After each collision, their directions change. So these electrons have no net flow in the particular direction and hence no current flows.

### Q. Why are alloys constantan and manganin used to make standard resistors?

Ans:

The resistance of a conductor also varies with temperature as

${{R}_{\theta }}$ = R0 [1 +  $\alpha \theta$ ] where $\alpha$ is the temperature coefficient of the resistance of the wire.

The alloys constantan and manganin are used to make standard resistors because the value of their temperature coefficient of resistance ( $\alpha$) is negligible. i.e. with change in temperature the resistance of such substance has negligible effect.

### Q. Give an example of non-ohmic conductor and present its current voltage characteristics graph.

Ans:

A conductor which does not obey Ohm’s law is known as non-ohmic conductor. For example: electrolyte, junction diode etc.

If a graph is plotted between current and voltage for non ohmic conductor, a straight line passing through origin will not be obtained. I-V graph of a junction diode (a non- ohmic conductor) is shown in figure.

### Q. How can a galvanometer be converted into ammeter? Explain.

Ans:

A galvanometer can be converted into ammeter by connecting a suitable shunt (a low value resistance) in parallel with the galvanometer. If the galvanometer has resistance ‘G’ and shows full scale deflection when an amount of current Ig flows through it. The required value of shunt to convert the galvanometer into an ammeter to measure current upto I ampere is

S = $\frac{{{I}_{g}}}{I-{{I}_{g}}}$.G

### Q. How can a galvanometer be converted into voltmeter? Explain.

Ans:

A galvanometer can be converted into voltmeter by connecting a high value resistance in series with the galvanometer. If the galvanometer has resistance ‘G’ and shows full scale deflection when an amount of current Ig flows through it. The required value of high value resistance to convert the galvanometer into a voltmeter to measure potential difference upto V volt is

R = $\frac{V}{{{I}_{g}}}$– G

Also Read: D.C. Circuit Notes Class 12