Unit 11: Electric Circuits
Weightage : 15–18%
The circuit shown has a battery of negligible internal resistance, resistors, and a switch. There are voltmeters, which measure the potential differences \(V_{1}\) and \(V_{2}\), and ammeters \(A_{1}\), \(A_{2}\), and \(A_{3}\), which measure the currents \(I_{1}\), \(I_{2}\), and \(I_{3}\), respectively. The switch is initially in the closed position. With the switch still closed, which of the following relationships are true? (Select two answers.)
Correct Answers: \( \boxed{\mathrm{A\ and\ B}} \)
Choice (A): Applying Kirchhoff’s Junction Rule at the node above ammeter \(A_{2}\), the total current entering the junction equals the total current leaving:
\( I_{1}+I_{2}-I_{3}=0 \)
Therefore, (A) is correct.
Choice (B): Applying Kirchhoff’s Loop Rule to the right-hand loop containing the battery, the sum of the potential changes around the loop is zero:
\( \Delta V_{B}-V_{1}-V_{2}=0 \)
Therefore, (B) is correct.
Choice (C): The voltmeters measure potential differences across different branches. Solving the circuit shows that \(V_{1}=V_{2}\), so \(V_{1}>V_{2}\) is false.
Choice (D): Ammeter \(A_{3}\) measures the total current supplied by the battery, whereas ammeter \(A_{2}\) measures only one branch current. Thus,
\( I_{3}>I_{2} \)
so \(I_{2}=I_{3}\) is false.
Therefore, the two correct answers are (A) and (B).
A battery with emf \( \varepsilon \) and internal resistance of \(30\,\Omega\) is being recharged by connecting it to an outlet with a potential difference of \(120\,\mathrm{V}\), as shown above. While it is being recharged, \(3\,\mathrm{A}\) flows through the battery.
Determine the emf of the battery.
Correct Answer: \( \boxed{\mathrm{D}} \)
Since the battery is being recharged, the external \(120\,\mathrm{V}\) source drives current into the battery. Applying Kirchhoff’s Loop Rule, the algebraic sum of the potential changes around the loop must be zero.
Traversing the loop in the direction of the current gives
\(120-\varepsilon-Ir=0\)
Substitute \(I=3\,\mathrm{A}\) and \(r=30\,\Omega\):
\(120-\varepsilon-(3)(30)=0\)
\(120-\varepsilon-90=0\)
\(\varepsilon=30\,\mathrm{V}\)
Therefore, the emf of the battery is \( \boxed{30\,\mathrm{V}} \).
Answer: (D)
Four identical batteries of negligible resistance are connected to resistors as shown. A voltmeter is connected to the points indicated by the dots in each circuit. Which of the following correctly ranks the potential difference measured by the voltmeter?
Correct Answer: \( \boxed{\mathrm{A}} \)
Apply Kirchhoff’s Loop Rule, which states that the algebraic sum of all potential differences around any closed loop is zero:
\( \sum \Delta V = 0 \)
Circuit A: The voltmeter measures the potential difference across the only resistor in the loop. Since there is a single resistor connected to the battery, the resistor must have the full battery voltage across it:
\( \Delta V_A = V_{\text{battery}} \)
Circuit B: Although there are two resistors, the resistor between the voltmeter terminals is connected directly across the battery. Therefore, it also has the full battery voltage across it:
\( \Delta V_B = V_{\text{battery}} \)
Circuit C: The measured resistor shares the battery voltage with another resistor in the same loop. Consequently, the battery voltage is divided between the resistors, so the voltmeter reads less than the full battery voltage:
\( \Delta V_C < V_{\text{battery}} \)
Therefore,
\( \boxed{\Delta V_A=\Delta V_B>\Delta V_C} \)
Answer: (A)
