[**NCERT Exercise**]

Two conducting wires of the same material and of equal lengths and equal diameters are first connected in series and then parallel in a circuit across the same potential difference. The ratio of heat produced in series and parallel combinations would be

Two conducting wires of the same material and of equal lengths and equal diameters are first connected in series and then parallel in a circuit across the same potential difference. The ratio of heat produced in series and parallel combinations would be –

(a) 1:2

(b) 2:1

(c) 1:4

(d) 4:1

Ans (c) 1:4

(c) The Joule heating is given by, H = i^{2}Rt

Let, R be the resistance of the two wires.

The equivalent resistance of the series connection is RS= R + R = 2R

If V is the applied potential difference, then it is the voltage across the
equivalent resistance.

The heat dissipated in time t is,

The equivalent resistance of the parallel connection is RP=

V is the applied potential difference across this RP.

The heat dissipated in time t is,

So, the ratio of heat produced is,

Note: H α R also H α i^{2} and H
α t. In this question, t is same for both the circuit.
But the current through the equivalent resistance of both the circuit is
different. We could have solved the question directly using H
α R if in case the
current was also same. As we know the voltage and resistance of the circuits, we
have calculated i in terms of voltage and resistance and used in the equation H
= i^{2}Rt to find the ratio.

[**NCERT Exercise**]

How is a voltmeter connected in the circuit to measure the potential difference between two points?

To measure the potential difference between two points, a voltmeter should be connected in parallel to the points.

[**NCERT Exercise**]

A copper wire has diameter 0.5 mm and resistivity of 1.6 � 10^{−8}Ω
m. What will be the length of this wire to make its resistance 10 Ω? How much
does the resistance change if the diameter is doubled?

A copper wire has diameter 0.5 mm and resistivity of 1.6 � 10^{−8}Ω
m. What will be the length of this wire to make its resistance 10 Ω? How much
does the resistance change if the diameter is doubled?

Answer

Area of cross-section of the wire, A =π (d/2) ^{2}

Diameter= 0.5 mm = 0.0005 m

Resistance, R = 10 Ω

We know that

Therefore, the length of the wire is 122.7 m and the new resistance is 2.5 Ω.

[**NCERT Exercise**]
The values of current I flowing in a given resistor for the corresponding
values of potential difference V across the resistor are given below −

I (amperes ) | 0.5 | 1.0 | 2.0 | 3.0 | 4.0 |

V (volts) | 1.6 | 3.4 | 6.7 | 3.0 | 13.2 |

Plot a graph between V and I and calculate the resistance of that resistor.

The plot between voltage and current is called IV characteristic. The voltage is plotted on x-axis and current is plotted on y-axis. The values of the current for different values of the voltage are shown in the given table.

V (volts) | 1.6 | 3.4 | 6.7 | 3.0 | 13.2 |

I (amperes ) | 0.5 | 1.0 | 2.0 | 3.0 | 4.0 |

The IV characteristic of the given resistor is plotted in the following figure.

The slope of the line gives the value of resistance (R) as,

Slope = 1/R = BC/AC = 2/6.8

R= 6.8/2 = 3.4 Ω

Therefore, the resistance of the resistor is 3.4 Ω.