**Problem 1:**

Find the resistance of a copper wire of 0.75 km long and having a cross sectional area of 0.01 cm

^{2}. (Take ρ = 1.72 x 10^{-8}ohm-m).
Given data

Length of wire, l = 0.75 km = 0.75 x 10

^{3}m
Cross sectional area, a = 0.01 cm

^{2}= 0.01 x 10^{-4}m^{2}
Specific resistance, ρ = 1.72 x 10

^{-8}ohm-m**Solution:**

Resistance, R = ρl/a = 1.72 x 10

^{-8}x 0.75 x 10^{3}/0.01 x 10^{-4}
= 2.9 Ω

**Problem 2:**

Find the cross sectional area of a aluminum wire of 700m long and having a resistance of 0.24 ohm (Take ρ = 2.83 x 10

^{-8}ohm-m)
Given data

Length of wire, l = 700 m

Resistance, R = 0.24 ohm

Specific resistance, ρ = 2.83 x 10

^{-8}ohm-m**Solution:**

Resistance, R = ρl/a

Cross sectional area, a = ρl/R = 2.83 x 10

^{-8}x 700/0.24 = 1.81 x 10^{-5}/0.24
a = 8.254 x 10

^{-5}m^{2}**Problem 3:**

Find the resistance of a copper wire 1 km long and 0.5 cm diameter given the specific resistance of copper as 1.7 x 10

^{-8}Ω-m
Given data

Length of wire in meter, l = 1 km = 1000 m

Specific resistance, ρ = 1.7 x 10

^{-8}Ω-m**Solution:**

Resistance, R = ρl/a = 1.7 x 10

^{-8}x 1000/ 0.785 x 2.5 x 10^{-5}
R = 0.866 ohm

**Problem 4:**

An electric iron is rated 1000 W and 240 V. Find the current drawn and resistance of the heating element.

Given data,

Power, P = V

^{2}/R
Resistance, R = V

^{2}/P = (240)^{2}/1000 = 57.6 Ohms
Current, I = V/R = 240/57.6 = 4.17 amps

**Problem 5:**

Calculate the current and resistance of a 50 W, 100 V electric bulb.

Given Data

Power, P = 50 watts

Voltage, V = 100 volts

**Solution:**

Power, P = VI watts

Current, I = P/V = 50/100 = 0.5 amps

Resistance, R = V/I = 100/0.5 = 200 Ohms

**Problem 6:**

Calculate the power rating of a heater coil when used on 200V supply taking 4 amps

**Solution:**

Power, P = VI watts = 200 x 4 = 800 watts

**Problem 7:**

A filament lamp connected to a 230V DC supply draws 300 mA. Find the power absorbed by the lamp.

**Solution:**

Power P = VI watts = 230 x 300 x 10

^{-3}= 69 watts**Problem 8:**

A lamp having a resistance of 500 Ω works on 220 supply system, Determine the energy consumed by operating 30 days at the rate of 4 hours a day.

**Solution:**

Energy = Power x Energy watt-sec

Power, P = V

^{2}/R = 220^{2}/500 = 96.8 watts
Total hours = 30 x 4 = 120

Energy = 96.8 x 120 = 11616 hr = 11.616 Kwhr

**Problem 9:**

A building has following loads : twenty 100 watt lamps operated 4 hours daily and thirty 60 watt lamps operated 3 hours daily, all connected to a 230 volt source. Calculate (a) the total power (b) the total current (c) the monthly consumption electrical energy (d) the monthly electrical energy charges at 35 paise per unit.

**Solution:**

(a) Total power, P = 20 x 100 + 30 x 60 = 3800 watts = 3.8 K

(b) Total current, I = P/V = 3800/230 = 16.5 amps

(c) Energy consumption daily = 20 x 100 x 4 + 30 x 60 x 3 = 13,400 watt-hours

Monthly consumption = 13400 x 30 Whr = 13.4 x 30 Kwh

(d) The monthly electrical energy charges at 35 paise per unit = 13.4 x 30 x 0.35 = Rs 140.70

## 0 on: "DC Circuits Problems and Solutions"