CURRENT BYU 21

We have a circuit with an ideal battery ($E$) and two resistors $R_1$ and $R_2$ in series. The first voltmeter ($R_V = 1.0 \text{ k}\Omega$) gives these readings:

• Reading across $R_1$: $V_1 = 1.8 \text{ V}$
• Reading across $R_2$: $V_2 = 1.8 \text{ V}$
• Reading across Battery (A-C): $V_{battery} = 4.5 \text{ V}$

Deductions:
1. Since the battery is ideal, its reading is the EMF: $E = 4.5 \text{ V}$.
2. Since $V_1 = V_2$, the resistors must be identical: $R_1 = R_2 = R$.

When the voltmeter is connected across $R_1$ (A-B), it is in parallel with $R_1$. The circuit becomes: a parallel combination of $(R \parallel R_V)$ in series with $R_2 (=R)$.
Voltage across parallel combination: $V_p = 1.8 \text{ V}$.
Voltage across $R_2$ (series component): $V_s = E – V_p = 4.5 – 1.8 = 2.7 \text{ V}$.

The current in the circuit is determined by the series resistor $R_2$: $$ I = \frac{V_s}{R} = \frac{2.7}{R} $$ This same current flows into the parallel combination $(R \parallel R_V)$: $$ V_p = I \cdot \left( \frac{R R_V}{R + R_V} \right) $$ $$ 1.8 = \frac{2.7}{R} \cdot \frac{R R_V}{R + R_V} $$ $$ 1.8 = \frac{2.7 R_V}{R + R_V} $$ $$ 1.8 (R + R_V) = 2.7 R_V $$ $$ 1.8 R = 0.9 R_V \implies R_V = 2R \implies R = \frac{R_V}{2} $$ Given $R_V = 1000 \, \Omega$, we get $R = 500 \, \Omega$.

We now use a new voltmeter with $R’_V = 2.0 \text{ k}\Omega = 2000 \, \Omega$.
Reading across A-C: Remains the battery EMF, $4.5 \text{ V}$.

Reading across A-B (or B-C): The setup is now: $(R \parallel R’_V)$ in series with $R$.
Equivalent resistance of the parallel part ($R_p$): $$ R_p = \frac{R R’_V}{R + R’_V} = \frac{500 \times 2000}{500 + 2000} = \frac{1000000}{2500} = 400 \, \Omega $$ Total circuit resistance: $$ R_{total} = R_p + R = 400 + 500 = 900 \, \Omega $$ The voltmeter reading is the potential drop across $R_p$: $$ V’_{reading} = E \cdot \frac{R_p}{R_{total}} = 4.5 \cdot \frac{400}{900} $$ $$ V’_{reading} = 4.5 \cdot \frac{4}{9} = 0.5 \cdot 4 = 2.0 \text{ V} $$