Question

# An 18.0 $\Omega$, 9.00 $\Omega$, and 6.00 $\Omega$ resistor are connected in series with an emf source. The current in the 9.00 $\Omega$ resistor is measured to be 4.00 A. Find the potential difference across the emf source.

Solution

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Answered 2 years ago
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$\textbf{Given:}$

\begin{align*} I_{\mathrm{total}} &= 4.00 \mathrm{\: A} \\ R_{18} &= 18.0 \Omega \\ R_{9} &= 9.00 \Omega \\ R_{6} &= 6.00 \Omega \\ R_{\mathrm{eq}} &= 33 \Omega \\ \end{align*}

To compute for the potential difference across the emf source, we must first know the total current and the equivalent resistance of the circuit. The equivalent resistance was computed in the previous part of this exercise. Since the circuit is in series, the total current of the circuit is the current in the 9.00 $\Omega$ resistor, which is 4.00 A.

With the equivalent resistance and total current known, we can then use Ohm's law to compute for the potential difference across the emf source:

\begin{align*} \Delta V &= I_{\mathrm{total}} \cdot R_{\mathrm{eq}} \\ &= 4.00 \mathrm{\: A} \cdot 33 \Omega\\ \Delta V &= \boxed{132 \mathrm{\: V}} \end{align*}

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