SSE JE Electrical Engineering 29 Oct 2020 Shift 2 Papers

Question : 135 of 200

Marks: +1, -0

A supply of 120 V is applied to three lamps connected in parallel. The power ratings of the lamps are 60 W, 40 W, and 100 W. Determine the total resistance and total current.

R = 42 Ω, I = 2.86 A
R = 52 Ω, I = 1.46 A
R = 72 Ω, I = 2.46 A
R = 72 Ω, I = 1.67 A

Validate

Solution:

Supply Voltage(V) = 120 V

\P _{ \B 1}=60 \W

\P_{ \B 2}=40 \W

\P _{ \B 3}=100 \W

From the above concept,

\R _{ \B 1}={ \V ^{2}}/{ \P _{ \B 1}}={120^{2}}{60}=240 \Ω

\R _{ \B 2}={ \V ^{2}}/{ \P _{ \B 2}}={120^{2}}{40}=360 \Ω

\R _{ \B 3}={ \V ^{2}}/{ \P _{ \B 3}}={120^{2}}{100}=144 \Ω

Therefore,

{1}/{\R}={1}/{\R_{\B 1}}+{1}/{\R_{\B2}}+{1}/{\R_{\B 3}}

={1}/{240}+{1}/{360}+{1}/{144}

On solving the value of

\R

\R=72 \Ω

And total Current,

\I ={ \V }/{ \R }={120}/{72}=1.67 \A

Go to Question:

Prev Question

Next Question