A galvanometer has a resistance of \(99~\Omega\) and deflects full scale when a current of \(10^{-3}~\mathrm{A}\) passes through it. In order to convert this galvanometer into an ammeter with a full-scale deflection of \(0.1~\mathrm{A}\), one should connect a resistance of
Correct Answer: \( \boxed{\mathrm{D}} \)
To convert a galvanometer into an ammeter, a low-resistance shunt is connected in parallel with the galvanometer so that most of the current bypasses the coil.
The galvanometer reaches full-scale deflection at
\( I_g=10^{-3}~\mathrm{A} \)
with resistance
\( R_g=99~\Omega \)
The voltage across the galvanometer at full scale is
\( V=I_gR_g=(10^{-3})(99)=0.099~\mathrm{V} \)
For a full-scale ammeter reading of \(0.1~\mathrm{A}\), the shunt current is
\( I_s=0.1-10^{-3}=0.099~\mathrm{A} \)
Since the shunt is in parallel with the galvanometer, it has the same voltage:
\( R_s=\dfrac{V}{I_s}=\dfrac{0.099}{0.099}=1~\Omega \)
Therefore, a \(1~\Omega\) resistor must be connected in parallel with the galvanometer.
Hence, the correct answer is (D).
Time \(t\) is the time it takes the current of an LR circuit with an inductor of inductance \(L\) and a resistor of resistance \(R\) to reach half of its maximum value. What is the new time if the original inductor is replaced with an inductor of inductance \(2L\)?
Correct Answer: \( \boxed{\mathrm{B}} \)
For an LR circuit, the current varies as
\( I=I_{\max}\left(1-e^{-t/\tau}\right) \)
where the time constant is
\( \tau=\dfrac{L}{R} \)
To reach one-half of the maximum current,
\( \dfrac{1}{2}=1-e^{-t/\tau} \)
which gives
\( t=\tau\ln 2=\dfrac{L}{R}\ln 2 \)
Thus, the time is directly proportional to the inductance:
\( t\propto L \)
If the inductance is doubled from \(L\) to \(2L\), the new time becomes
\( t_{\mathrm{new}}=2t \)
Hence, the correct answer is (B).
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)
