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Question
An electric bulb of resistance 20 Ω and a resistance wire of 4 Ω are connected in series with a 6 V battery.
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.
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Solution
(a) Total resistance, R = 20 Ω + 4 Ω = 24 Ω
(b) Current through the circuit,
I = V/R
I = 6/24
I = 0.25 A
(c) Potential difference across the electric bulb,
Vbulb = IRbulb
`=0.25xx20`
= 5 V
(d) Potential difference across the resistance wire,
Vwire = IRwire
`= 0.25xx4`
= 1 V
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