**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

**Problem 10:**

In the circuit shown in fig., Find the current , the voltage
across each resistor and the power dissipated in each resistor.

Solution:

Total resistance, R = R

_{1}+ R_{2}+ R_{3}= 6 + 9 + 10 = 25 Ω
I = V/R = 50/25 = 2 amps

Voltage drop across 6 Ω resistor, V

_{1}= IR_{1}= 2 x 6 = 12 volts
Voltage drop across 9 Ω resistor, V

_{2}= IR_{2}= 2 x 9 = 18 volts
Voltage drop across 10 Ω resistor, V

_{3}= IR_{3}= 2 x 10 = 20 volts
Power dissipated in 6 Ω resistor, P

_{1}= I^{2}R_{1}= 4 x 6 = 24 watts
Power dissipated in 9 Ω resistor, P

_{2}= I^{2}R_{2}= 4 x 9 = 36 watts
Power dissipated in 10 Ω resistor, P

_{3}= I^{2}R_{3}= 4 x 10 = 40 watts**Problem 11:**

Two resistors resistance 6 Ω and 12 Ω are connected in
series. Find the equivalent Also find the equivalent resistance when they are
connected in parallel.

Given data

Case 1: Series combination of 6 Ω and 12 Ω
resistors.

R

_{1}= 6 Ω, R_{2}= 12 Ω
Solution:

Equivalent resistance, R = R

_{1 }+ R_{2 }= 18 Ω
Case 2: Parallel combination of 6 Ω and 12 Ω resistors.

R1 = 6 Ω, R2 = 12 Ω

Equivalent resistance, 1/R = 1/ R

_{1 }+ 1/R_{2}
R = R

_{1}R_{2}/(R_{1}+ R_{2}) = 6 x 12/(6 + 12) = 72/18 = 4 Ω**Problem 12:**

In the circuit shown in figure , find
the total resistance and current flowing through each branch.

Given data, R

_{1}= 5 Ω, R_{2}= 10 Ω, R_{3}= 20 Ω
Solution:

Equivalent resistance, 1/R = 1/ R

_{1 }+ 1/R_{2}+ 1/R_{3 }= 0.2 + 0.1 + 0.05 = 0.35
Total resistance, R = 2.86 Ω

The current flowing through 5 Ω resistor, I

_{1}= V/R_{1}= 25/5 = 5 amps.
The current flowing through 10 Ω resistor, I

_{2}= V/R_{2}= 25/10 = 2.5 amps
The current flowing through 20 Ω resistor, I

_{3}= V/R_{3}= 25/20 = 1.25 amps**Problem 13:**

Resistors of values 2, 3, 4, and 5 ohm are connected in
parallel. If the total power absorbed by all the resistors is 200w, find the
voltage applied to the circuit.

Power, P = 200 watts

R

_{1}= 2 Ω, R_{2}= 3 Ω, R_{3}= 4 Ω, R_{4}= 5 Ω
Solution:

Total Resistance, 1/R = 1/ R

_{1 }+ 1/R_{2}+ 1/R_{3}+ 1/R_{4}= 0.5 + 0.33 +0.25 + 0.2
1/R = 1.28, R = 1/1.28 = 0.78 ohm

Power, P = V2/R

V

^{2}= PX R
V = √(P X R) = √
(200 X 0.78) = 12.5 volts

**Problem 14:**

Two resistors are connected in
parallel across a 200V mains take a total current of 10amps. The power
dissipated in one of the resistors is 1200. watts. Find the value of each
resistor.

Given data:

Voltage, V = 200 volts

Current, I = 10 amps

Power dissipated in R

_{1}of the resistors P_{1}= 1200 watts
Solution:

I

_{1}= P_{1}/V = 1200/ 200 = 6 amps.
Resistors, R

_{1}= V/ I_{1}= 200/6 = 33.33 Ω
Total current, I = I

_{1 }+ I_{2}
10 = 6 + I

_{2}
I

_{2}= 4 amps
Resistors, R

_{2}= V/ I_{2}= 200/4 = 50 Ω