Unit 11: Electric Circuits
Weightage : 15–18%
Each of the resistors shown in the circuit below has a resistance of \(200\,\Omega\). The emf of the ideal battery is \(24\,\mathrm{V}\).
What is the ratio of the power dissipated by \(R_{1}\) to the power dissipated by \(R_{4}\)?
Correct Answer: \( \boxed{\mathrm{E}} \)
Resistor \(R_{1}\) is in series with a parallel combination of \(R_{2}\), \(R_{3}\), and \(R_{4}\). Since all three parallel resistors have the same resistance (\(200\,\Omega\)), the current divides equally among them.
Therefore,
\( I_{4}=\dfrac{I_{1}}{3} \)
The power dissipated by a resistor is
\( P=I^{2}R \)
Since \(R_{1}=R_{4}=200\,\Omega\),
$ \frac{P_{1}}{P_{4}} = \frac{I_{1}^{2}R_{1}} {I_{4}^{2}R_{4}} = \frac{I_{1}^{2}} {\left(\dfrac{I_{1}}{3}\right)^{2}} =9$
Thus, resistor \(R_{1}\) dissipates nine times as much power as resistor \(R_{4}\).
Answer: (E)
A hair dryer is rated as \(1200\,\mathrm{W},\ 120\,\mathrm{V}\). Its effective internal resistance is
Correct Answer: \( \boxed{\mathrm{C}} \)
The power rating of an electrical device is related to its resistance by
\( P=\dfrac{V^2}{R} \)
Rearranging for resistance,
\( R=\dfrac{V^2}{P} \)
Substituting the given values,
\( R=\dfrac{(120\,\mathrm{V})^2}{1200\,\mathrm{W}}=\dfrac{14400}{1200}=12\,\Omega \)
Therefore, the effective internal resistance of the hair dryer is \( \boxed{12\,\Omega} \).
Answer: (C)
A \(120\,\mathrm{V}\) source is connected to three resistors, as shown in the diagram. The resistance of \(R_3\) is twice the resistance of \(R_2\), which is twice the resistance of \(R_1\).
Which of the following ranks the power \(P_1\), \(P_2\), and \(P_3\) dissipated by each resistor?
Correct Answer: \( \boxed{\mathrm{E}} \)
The three resistors are connected in parallel, so each resistor has the same potential difference across it:
\(V=120\,\mathrm{V}\)
The power dissipated by a resistor is given by
\(P=\dfrac{V^2}{R}\)
Since the voltage is the same for all three resistors, the power is inversely proportional to the resistance.
Given
\(R_2=2R_1\)
\(R_3=2R_2=4R_1\)
Therefore,
\(P_1=\dfrac{V^2}{R_1}\), \(P_2=\dfrac{V^2}{2R_1}=\dfrac{P_1}{2}\), \(P_3=\dfrac{V^2}{4R_1}=\dfrac{P_1}{4}\)
Thus, the ranking of the powers is
\(P_1>P_2>P_3\)
The physical distance of a resistor from the battery does not affect the voltage across it in a parallel circuit.
Answer: (E)
