Question

unlike the idealized voltmeter, a real voltmeter has a resistance that is not infinitely large. part a a voltmeter with resistance $r_v$ is connected across the terminals of a battery of emf $\mathcal{e}$ and internal resistance $r$. find the potential difference $v_{meter}$ measured by the voltmeter. view available hint(s) $v_{meter}=$ submit part b complete previous part(s)

Explanation:

Step1: Find total resistance

The volt - meter with resistance $R_V$ is in parallel with the battery's terminals. The total resistance $R_{total}$ of the circuit is the equivalent resistance of $R_V$ and the internal resistance $r$ in parallel. Using the formula for parallel resistors, the equivalent resistance $R_{eq}=\frac{R_V\times r}{R_V + r}$, and the total resistance $R_{total}=\frac{R_V\times r}{R_V + r}$.

Step2: Find the current in the circuit

According to Ohm's law $I=\frac{\mathcal{E}}{R_{total}}$, substituting $R_{total}$, we get $I=\frac{\mathcal{E}(R_V + r)}{R_V\times r}$.

Step3: Find the potential difference across the volt - meter

The potential difference across the volt - meter $V_{meter}$ is given by $V_{meter}=I\times\frac{R_V\times r}{R_V + r}$. Substituting $I$ into this equation, we have $V_{meter}=\frac{\mathcal{E}R_V}{R_V + r}$.

Answer:

$V_{meter}=\frac{\mathcal{E}R_V}{R_V + r}$