Question

A cable of resistance 12ohm carries electric power from a generator producing 250 kW at 10,000 V. Calculate (i) the current in the cable, (ii) the power lost in the cable during transmission, and (iii) the p.d. across the ends of the cable.

Answer 1

R=12 ohm

P=250kW = 250000w

V=1000v

I=p/v = 250000/10000=25A

P=I^2 R= 25×25×12 = 7500 w =7.5kW

V=IR = 25 × 12 = 300 V

Answer 2

The current in the table is 25 A

The Power lost in the cable is 7500 W

The potential difference across the table is 300 volts

Given:  

Resistance (R) = 12 ohm

Power (P) = 250 KW = 250000 W

Voltage (V) = 10000V

Solution:

According to Formula of Power:

$\mathrm{P}=\mathrm{V} \times I$ ………. (1)  

Where, P = power, V= voltage, I= current

From equation 1,  

Given that,

$\begin{aligned} 250000 &=10000 \times(I) \\\\ I &=\frac{250000}{10000} \end{aligned}I = 25 A$

Hence Current in circuit = 25 A

Given that, Resistance (R) = 12 ohm

Now, according to Kirchhoff’s Law, power loss (p)

$\begin{array}{l}{P=I^{2} \times R} \\ {P=(25)^{2} \times 12}\end{array}$

P= 7500 W

The power lost in the cable during transmission = 7500W

(iii)   Now,  

potential difference (Voltage) across the end of cable -

From Ohms law,

$\begin{aligned} V &=R \times I \\ V &=12 \times 25 \end{aligned}$

V = 300 volts