Question

A battery of 9 V is connected in series with resistors of 0.2 Ω, 0.3 Ω, 0.4 Ω, 0.5 Ω and 0.12 Ω. How much current would flow through the 12 Ω resistor?

Answer 1

When five resistors are connected in series, the total resistance is given by:
R = R₁+ R₂+ R₃+ R₄+ R₅
Here,
R₁= 0.2 $\mathrm{\Omega }$,
R₂= 0.3 $\mathrm{\Omega }$,
R₃= 0.4 $\mathrm{\Omega }$,
R₄= 0.5 $\mathrm{\Omega }$,
R₅= 0.12 $\mathrm{\Omega }$

Therefore:
R = 0.2 Ω + 0.3 Ω + 0.4 Ω + 0.5 Ω + 12 Ω
R = 13.4 Ω
Total resistance of the circuit = 13.4 Ω
The current flowing through this series combination is given by I = V / R.
or I = 9 / 13.4 = 0.67 A
Now since the resistors are connected in series, the current flowing through each resistance is the same. Hence, the current through the 12 Ω resistor is equal to 0.67 A.