Question

In the circuit of Fig. E25.30, the 5.0 Ω resistor is removed and replaced by a resistor of unknown resistance R. When this is done, an ideal voltmeter connected across the points band creads 1.9 V. Find (a) the current in the circuit and (b) the resistance R. (c) Graph the potential rises and drops in this circuit (see Fig. 25.20).

Short answer

1. The current in the circuit is 0.21A.
2. The Resistance R is 26.1$\Omega$.
3. The required graph is shown.

Step-by-step solution

  (a) Determination of the current in the circuit.

The sum of the potential across the loop is zero.

The current flows from + terminal to – terminal, the current will flow the counter clockwise direction as determined by the 16-V battery. The voltmeter reading is across the 9.0 Ω resistor.

${\mathrm{V}}_{\mathrm{bc}}=1.9\mathrm{V}$

Therefore, the current according to Ohm’s Law,

$$\begin{gathered} l=\frac{V_{ba}}{R_{bc}} \\ =\frac{1.9V}{9.0\Omega } \\ =0.21\mathrm{A} \\ \mathrm{Thus},\mathrm{the}\mathrm{current}\mathrm{in}\mathrm{the}\mathrm{circuit}\mathrm{is}0.21\mathrm{A}. \end{gathered}$$

(b) Determination of the resistance in the circuit.

Write the potential equation of the circuit as,

$$\begin{gathered} 16.0\mathrm{V}-8.0\mathrm{V}=(1.6\mathrm{\Omega }+9.0\mathrm{\Omega }+1.4\mathrm{\Omega }+\mathrm{R})(0.21\mathrm{A}) \\ \mathrm{R}=\frac{5.48\mathrm{V}}{0.21\mathrm{A}} \\ =26.1\mathrm{\Omega } \end{gathered}$$

 Graph for the potential rises and drops in this circuit.

The graph is sketched by below.When the 5.0-Ω resistor is replaced by the 26.1-Ω

resistor the current decreases to 0.21 A.