Home / AP Physics C E&M – 11.5 Steady-State Direct- Current Circuits with Batteries and Resistors Only- Exam Style questions- MCQs
Three resistors are connected to an ideal battery, as shown in the figure above. The battery has an emf \( \varepsilon \). Two of the resistors have known resistances \(R\) and \(2R\). The third resistor has unknown resistance \(X\). The current in two of the branches is shown. What is the value of the unknown resistance \(X\)?
Correct Answer: \( \boxed{\mathrm{C}} \)
Apply Kirchhoff’s Junction Rule at the central junction.
The current entering from the right is \(I\), while \(2I\) leaves along the lower-left branch. Therefore, the current through the vertical resistor must be
\(I_R = 2I – I = I.\)
The voltage drop across the vertical resistor is
\(V = IR.\)
The lower-right resistor \(X\) carries the remaining current \(2I\), so it has the same voltage:
\(2IX = IR.\)
Solving for \(X\),
\(X = \dfrac{R}{2}.\)
Therefore, the correct answer is (C).
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: \( \boxed{\mathrm{E}} \)
Resistors \(R_2\), \(R_3\), and \(R_4\) are identical and connected in parallel, so the current entering the parallel combination divides equally among the three branches.
If the current through \(R_1\) is \(I\), then the current through each of \(R_2\), \(R_3\), and \(R_4\) is
\( I_4=\dfrac{I}{3} \)
The power dissipated in a resistor is
\( P=I^2R \)
Since \(R_1=R_4=200~\Omega\),
\( \dfrac{P_1}{P_4}=\dfrac{I^2R}{\left(\dfrac{I}{3}\right)^2R}=\dfrac{I^2}{I^2/9}=9 \)
Therefore,
\( \boxed{\dfrac{P_1}{P_4}=9} \)
Hence, the correct answer is (E).
A \(120~\mathrm{V}\) source is connected to three resistors in parallel, 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 is correct about the equivalent resistance of this circuit?
Correct Answer: \( \boxed{\mathrm{D}} \)
For resistors connected in parallel, the equivalent resistance satisfies
\( \dfrac{1}{R_{\mathrm{eq}}}=\dfrac{1}{R_1}+\dfrac{1}{R_2}+\dfrac{1}{R_3} \)
Since each term on the right-hand side is positive, adding more parallel branches always decreases the equivalent resistance.
A key property of parallel circuits is that the equivalent resistance is always less than the smallest individual resistance.
Here, \(R_1\) is the smallest resistor because
\( R_2=2R_1,\qquad R_3=4R_1 \)
Therefore,
\( \boxed{R_{\mathrm{eq}}<R_1} \)
Hence, the correct answer is (D).