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Question
Answer the following question.
An electric lamp of resistance 20 Ω and a conductor of resistance 4 0 are connected to a 6 V battery as shown in the circuit. Calculate: 
(a) the total resistance of the circuit,
(b) the current through the circuit,
(c) the potential difference across the (i) electric lamp and (ii) conductor, and
(d) power of the lamp.
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Solution
Resistance of electric lamp = 20 Ω
Resistance of conductor = 4 Ω
Voltage battery = 6 V
(a) Total resistance of circuit = 20 Ω + 4 Ω = 24
(b) Resistance of conductor= 4 Ω
Voltage battery = 6 V
Apply Ohms law
6 V =` "I"` × 24 Ω
`"I" = (6"V")/(24Ω) = 0.25"A"`
Hence, current in the circuit is 0.25A
(c)
(i) Potential difference across the lamp
Vlamp = IR
Vlamp = 0.25 A × 20 Ω = 5 V
∴ Vlamp = 5 V
(ii) Potential difference across the conductor
`"V"_"conductor" = "IR"`
`"V"_"conductor" = 0.25 "A"xx4 Ω = 1 "V"`
`"V"_"conductor" = 1"V"`
(d) Power of lamp
`"I"^(2)"R" = (0.25)^2 xx 20 = 1.25 "W"`
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