WAEC Physics 2017 Theory — Question 28

Model answer

Resolving the two parallel branches (400Ω combined with the 800Ω voltmeter branch): Total resistance in the circuit where R1+R2=RT. 1/R1=1/400+1/400=2/400, so R1=200Ω. And 1/R2=1/800+1/800 [NB: the other is the resistance of the voltmeter], giving R2=400Ω (⇒Ω=400Ω). RT=(200+400)Ω=600Ω. From V=IR: VT=IT.RT. VT=6V; RT=600Ω ∴ IT=VT/RT=6/600=0.01A. V_voltmeter=I_voltmeter x R_voltmeter=IT x R2=0.01x400=4.0V.