Question

unlike the idealized voltmeter, a real voltmeter has a resistance that is not infinitely large. part b if $mathcal{e}=7.50 v$ and $r = 0.45 omega$, find the minimum value of the voltmeter resistance $r_v$ for which the voltmeter reading is within 1.0% of the emf of the battery. express your answer numerically (in ohms) to at least three significant digits. view available hint(s) $r_v= omega$ submit previous answers incorrect; try again; 2 attempts remaining

Explanation:

Step1: Determine the circuit - current formula

The emf of the battery is $\mathcal{E}$, the internal resistance is $r$, and the volt - meter resistance is $R_V$. The current in the circuit $I=\frac{\mathcal{E}}{r + R_{eq}}$, where $R_{eq}$ is the equivalent resistance of the circuit. Since the volt - meter is in parallel with the battery terminals, $R_{eq}=\frac{R_V\times R}{R_V + R}$ (assuming no other resistance in the circuit for the voltage - measurement part, and here we consider the case where the volt - meter is the only non - ideal element). The volt - meter reading $V = IR_{eq}=\frac{\mathcal{E}R_{eq}}{r + R_{eq}}$. We want $V\geq0.99\mathcal{E}$.

Step2: Substitute and simplify the inequality

Substituting $R_{eq}=\frac{R_V\times R}{R_V + R}$ (in our case, we can assume $R$ is effectively infinite as we are only interested in the effect of the volt - meter resistance) into the inequality $\frac{\mathcal{E}R_{eq}}{r + R_{eq}}\geq0.99\mathcal{E}$. Canceling out $\mathcal{E}$ on both sides, we get $\frac{R_V}{r + R_V}\geq0.99$. Cross - multiply: $R_V\geq0.99r+0.99R_V$.

Step3: Solve for $R_V$

Rearrange the inequality $R_V-0.99R_V\geq0.99r$. So, $0.01R_V\geq0.99r$. Then $R_V\geq\frac{0.99r}{0.01}=99r$.

Step4: Plug in the value of $r$

Given $r = 0.45\ \Omega$, then $R_V\geq99\times0.45=44.55\ \Omega$.

Answer:

$44.6\ \Omega$