An electric lamp whose resistance is 8 ohms and a conductor of 2 ohms resistance are
connected in series to a 5 volt battery. Calculate
A. The total resistance of the circuit
B. The current through the circuit
C. The potential difference across the electric lamp and the conductor
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Answer:
(a) In series combination,
R=R
Lamp
+R
conductor
=10Ω+2Ω=12Ω
I=
R
V
=
12Ω
6v
=0.5A
⇒V
Lamp
=IR=0.5×10=5V
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