Question

The figure shows three ammeters A, B and C The ammeter B reads 0.5 A . If all ammeters have negligible resistance calculate:

(a) reading of ammeters C and A.

(b) the total resistance of the circuit.

Answer 1

Ammeter B measures the current through the \(6\ \Omega\) branch.

(a) The potential difference across the parallel branches is:

\[V=IR\]

\[=0.5\times6\]

\[=3\ \mathrm{V}\]

Thus the current through the (3\ \Omega) branch, measured by C, is:

\[I_C=\frac{3}{3}\]

= 1 A

The total current, measured by A, is:

\[I_A=0.5+1\]

= 1.5 A

(b) The parallel pair has resistance:

\[R_p=\frac{6\times3}{6+3}\]

= 2 Ω

Adding the series \(2\ \Omega\) resistor gives:

\[R_{\mathrm{total}}=2+2\]

= 4 Ω