# Electricity Worksheet-14

Electricity Worksheet-14

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

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

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

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

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

1. 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 ?

1. 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) 1-6       1.6      3.4       6.7     10.2     13.2

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

1. Q = 96000 C, t = 1 h = 60 × 60 = 3600 s, V = 50 V

Heat produced, W = QV = 96000 C × 50V = 48 × 105 J.

1. R = 20 Ω, I = 5 A, t = 30 s

Heat produced W = I2 Rt = (5)2 × 20 × 30 = 15000 J.

1. Electric power

1. 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.

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

1. 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.

1. 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,