Electricity Worksheet14

Compute the heat generated while transferring 96000 coulombs of charge in one hour through a potential difference of 50 V.

An electric iron of resistance 20 takes a current of 5 A. Calculate the heat produced in 30 s.

What determines the rate at which energy is delivered by a current .

An electric motor takes 5 A from a 220 V line. Determine the power and energy consumed in 2 h.

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

A copper wire has a diameter of 0.5 mm and a resistivity of 1.6 × 10^{–6} ohm cm. How much of this wire would be required to make a 10 ohm coil ? How much does the resistance change if the diameter is doubled ?

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 (ampere) 0.5 1.0 2.0 3.0 4.0
V (volt) 16 1.6 3.4 6.7 10.2 13.2
Plot a graph between V and I and calculate the resistance of the resistor.
Answer Key:

Q = 96000 C, t = 1 h = 60 × 60 = 3600 s, V = 50 V
Heat produced, W = QV = 96000 C × 50V = 48 × 10^{5} J.

R = 20 Ω, I = 5 A, t = 30 s
Heat produced W = I^{2} Rt = (5)^{2} × 20 × 30 = 15000 J.

Electric power

I = 5A, V = 220V, t = 2 h = 2 × 60 × 60 = 7200 s
Power P = VI = 220 × 5 = 1100 W
Energy consumed, W = VI t = 220 × 5 × 7200 = 792000 J.

A voltmeter is always connected in parallel across the points between which the potential difference is to be determined.

Diameter of the wire, D = 0.5 mm = 0.5 × 10^{–3} m
resistivity of copper, ρ = 1.6 × 10^{–6} ohm cm = 1.6 × 10^{–8} ohm m
required resistance, R = 10 ohm
As
or
When D is doubled, R becomes 1/4 times.

V – I graph
For V = 4 V (i.e., 9 V – 5 V), I = 1.25 A (i.e., 2.65 A – 1.40 A). Therefore,