Question

A circuit consists of three resistors of 3Omega 4Omega and 6Omega in parallel and a fourth resistor of 4Omega in series.A battery of e.m.f.12V and internal resistance 6Omega is connected across the circuit.Find the total current in the circuit and terminal voltage across the battery.​

Answer 1

Answer:

Given: V=12V, Resistance of A=4Ω, Resistance of B=6Ω, Internal battery resistance =2Ω

(i) Total Resistance =4+6+2=12Ω

I (current in circuit)=

Totalresistance

e.m.f.

=

12

12

=1amp

(ii) Voltage drop Ir=1×2=2volt

Terminal Voltage = emf - Voltage drop

=12−2=10 volts

(iii) Potential difference across 6Ω=IR

=1×6=6V

(iv) Energy spent =I

2

Rt

=(1)

2

×4×60=240J

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