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.

Answer 1

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
V_lamp = IR
V_lamp  = 0.25 A × 20 Ω = 5 V
∴ V_lamp = 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"`