Question

Remaining Time: 1 hour, 05 minutes, 56 seconds. Question Completion Status: 4 5 6 9 1 3 7 8 10 11 15 2 12 14 13 An electrical circuit is connected to 24 mV voltage source. The total current (in units of A) in the circuit is 5 Ω 2502 I w 2002 | 1.6 mA 0.625 mA 1.6 A 0.53 mA L Moving to another question will save this response.

Answer 1

As 25$\Omega$ and 5$\Omega$ are in series, their equivalent resistance=25+5=30$\Omega$

This 30$\Omega$ and other two 30$\Omega$ resistances are in parallel. Their equivalent Rp is

1/Rp=(1/30)+(1/30)+(1/30)=3/30=1/10

Rp=10$\Omega$

This Rp and 20$\Omega$ are in series. Their equivalent =Rp+20=10+20=30$\Omega$

This 30$\Omega$ and remaining 30$\Omega$( connected in parallel with battery) are in parallel.

Hence total resistance of circuit R is

1/R=(1/30)+(1/30)=2/30=1/15

R=15$\Omega$

Total voltage=24mV

Total current I=V/R=24/15=1.6mA

First Option