Unit 11: Electric Circuits
Weightage : 15–18%
The diagram shows a circuit that contains a battery with a potential difference of \(V_{B}\) and negligible internal resistance; five resistors of identical resistance; three ammeters \(A_{1}\), \(A_{2}\), \(A_{3}\); and a voltmeter.
What will be the reading of the voltmeter?
Correct Answer: \( \boxed{\mathrm{C}} \)
From the previous question, the equivalent resistance of the circuit to the right of the battery is
\(R_{\mathrm{right}}=\dfrac{5R}{3}\).
Therefore, the current measured by ammeter \(A_{2}\) is
$ I=\frac{V_{B}}{\frac{5R}{3}}=\frac{3V_{B}}{5R}. $
This current first passes through the resistor immediately to the right of ammeter \(A_{2}\), so the voltage drop across that resistor is
$ \Delta V=IR=\left(\frac{3V_{B}}{5R}\right)R=\frac{3V_{B}}{5}. $
The voltmeter is connected across the remaining parallel section of the circuit, so it measures the remaining potential difference:
$ V_{\mathrm{meter}}=V_{B}-\frac{3V_{B}}{5}=\frac{2V_{B}}{5}. $
Thus, the voltmeter reads
$ \boxed{\frac{2}{5}V_{B}}. $
Answer: (C)
In the circuit above, with the switch closed, what will voltmeter \(V\) read? (\(A\) is an ideal ammeter.) Select two answers.
Correct Answers: \( \boxed{\mathrm{A\ and\ C}} \)
The voltmeter is connected across the same two circuit nodes as both \(R_{1}\) and \(R_{2}\). Components connected across the same pair of nodes are in parallel and therefore have the same potential difference.
Consequently, the voltmeter measures the voltage drop across both \(R_{1}\) and \(R_{2}\).
Thus,
\( V=V_{R_1}=V_{R_2} \)
An ideal ammeter has negligible resistance, so there is no potential difference across it:
\( V_A=0 \)
Therefore, the voltmeter does not read the voltage across the ammeter, and its reading is not zero because current flows through \(R_{1}\) and \(R_{2}\).
Hence, the two correct answers are (A) and (C).
The above graph shows current as a function of potential difference for two different filament lamps. If the two lamps are connected in parallel to a \(3.0\,\mathrm{V}\) battery, what is the total current supplied by the battery?
Correct Answer: \( \boxed{\mathrm{D}} \)
In a parallel circuit, each lamp has the full battery potential difference across it. Therefore, each lamp operates at \(3.0\,\mathrm{V}\).
From the graph at \(3.0\,\mathrm{V}\):
• Upper lamp: \(I \approx 1.0\,\mathrm{A}\)
• Lower lamp: \(I \approx 0.6\,\mathrm{A}\)
By Kirchhoff’s junction rule, the total current supplied by the battery is the sum of the currents through the two parallel branches:
\( I_{\mathrm{total}} = I_1 + I_2 \)
\( I_{\mathrm{total}} = 1.0 + 0.6 = 1.6\,\mathrm{A} \)
Therefore, the total current supplied by the battery is \(1.6\,\mathrm{A}\).
Therefore, the correct answer is (D).
