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,
V_bulb = IR_bulb
=0.25 $\times$ 20
= 5 V
(d) Potential difference across the resistance wire,
V_wire = IR_wire
= 0.25 $\times$ 4
= 1 V

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(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 =0.25×20 = 5 V (d) Potential difference across the resistance wire, Vwire = IRwire = 0.25×4 = 1 V

(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,
V_bulb = IR_bulb
=0.25 $\times$ 20
= 5 V
(d) Potential difference across the resistance wire,
V_wire = IR_wire
= 0.25 $\times$ 4
= 1 V