QQuestion
In the combinations of resistors shown below, calculate:
(a) the resistance across AB when switch S is open.
(b) the resistance across AB when switch S is closed.
ICSEClass 10PhysicsCurrent Electricity04 Mark2026 Hard
✓Answer
(a) When the switch is open, the middle connection does not join the two branches. Therefore, the circuit has two separate branches in parallel. In each branch, the 12 Ω and 6 Ω resistors are in series.
For the top branch:
R1 = 12 + 6 = 18 Ω
For the bottom branch:
R2 = 12 + 6 = 18 Ω
Since R1 and R2 are in parallel:
\frac{1}{\mathrm{R}}=\frac{1}{\mathrm{R_1}}+\frac{1}{\mathrm{R_2}}or, \frac{1}{\mathrm{R}}=\frac{1}{18}+\frac{1}{18}=\frac{2}{18}=\frac{1}{9}
Therefore, R = 9 Ω.
Hence, the resistance across AB when switch S is open is 9 Ω.
(b) When switch S is closed, the midpoints are connected. The circuit is rearranged into two parallel combinations connected in series.
For the left combination containing 12 Ω and 6 Ω:
\frac{1}{\mathrm{R_1}}=\frac{1}{12}+\frac{1}{6}=\frac{1}{12}+\frac{2}{12}=\frac{3}{12}=\frac{1}{4}Therefore, R1 = 4 Ω.
Similarly, for the right combination:
\frac{1}{\mathrm{R_2}}=\frac{1}{12}+\frac{1}{6}=\frac{1}{12}+\frac{2}{12}=\frac{3}{12}=\frac{1}{4}Therefore, R2 = 4 Ω.
These two combinations are in series:
R = R1 + R2 = 4 + 4 = 8 Ω.
Hence, the resistance across AB when switch S is closed is 8 Ω.
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