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.
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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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