5. An electric bulb of resistance 20 Q and a resistance wire of 42 are connected in series with a 6 V batter
Draw the circuit diagram and calculate :
(a) total resistance of the circuit.
(b) current through the circuit.
(C) potential difference across the electric bulb.
(d) potential difference across the resistance wire.
Three resistors are connected as shown in the diagram.
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Answer:
Explanation:
(a) Total resistance= R1+R2=20+42=62Ω
(b) V=IR
I=V/R
=6/62
=0.1A
(c) V=IR
=0.1x20
=2V
(d) V=0.1x42
=4.2V
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