The amount of heat (H) produced is given by the joule’s law of heating asH= Vlt
Where,
Current, I = 5 A
Time, t = 30 s
Voltage, V = Current x Resistance = 5 x 20 = 100 V

H= 100 x 5 x 30 = 1.5 x 10⁴ J.

Therefore, the amount of heat developed in the electric iron is 1.5 x 10⁴J.

The rate of consumption of electric energy in an electric appliance is called electric power. Hence, the rate at which energy is delivered by a current is the power of the appliance.

Power (P) is given by the expression,P = VI
Where,
Voltage,V = 220 V
Current, I = 5 A
P= 220 x 5 = 1100 W
Energy consumed by the motor = Pt
Where,
Time, t = 2 h = 2 x 60 x 60 = 7200 s
∴ P = 1100 x 7200 = 7.92 x 10⁶ J
Therefore, power of the motor = 1100 W
Energy consumed by the motor = 7.92 x 10⁶ J

A piece of wire of resistance R is cut into five equal parts. These parts are then connected in parallel. If the equivalent resistance of this combination is R’, then the ratio R/R’ is -
(a) 1/25
(b) 1/5
(c) 5
(d) 25

Ans (d) 25

Which of the following terms does not represent electrical power in a circuit?
(a) I²R
(b) IR²
(c) VI
(d) V²/R

Ans (b) IR²

An electric bulb is rated 220 V and 100 W. When it is operated on 110 V, the power consumed will be –
(a) 100 W
(b) 75 W
(c) 50 W
(d) 25 W

Ans (d) 25 W

(d)Energy consumed by an appliance is given by the expression,





Where,

Power rating, P = 100 W

Voltage, V = 220 V

Resistance, R =

The resistance of the bulb remains constant if the supply voltage is reduced to 110 V. If the bulb is operated on 110 V, then the energy consumed by it is given by the expression for power as



Therefore, the power consumed will be 25 W

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 = i²Rt

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 i² 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 = i²Rt to find the ratio.

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