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Tuesday, 16 July 2019

DC Circuits Problems and Solutions

Problem 1:

Find the resistance of a copper wire of 0.75 km long and having a cross sectional area of 0.01 cm2. (Take ρ = 1.72 x 10-8 ohm-m).

Given data
Length of wire, l = 0.75 km = 0.75 x 103 m
Cross sectional area, a = 0.01 cm2 = 0.01 x 10-4 m2
Specific resistance, ρ = 1.72 x 10-8 ohm-m

Solution:

Resistance, R = ρl/a = 1.72 x 10-8 x 0.75 x 103/0.01 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 m2

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 = V2/R
Resistance, R = V2/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 = V2/R = 2202/500 = 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


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