Question

A box P and a coil Q are connected in series with an ac source of variable frequency. The emf of the source is constant at $10\text{V}$. Box P contains a capacitance of $1\mu F$ in series with a resistance of $32\Omega$. Coil Q has a self inductance of 4.9 mH and a resistance of $68\Omega$ in series. The frequency is adjusted so that maximum current flows in P and Q.
The voltage across P is

• A. $12\text{V}$
• B. $7.7\text{V}$
• C. $10\text{V}$
• D. $24\text{V}$

Answer 1

The correct option is B $7.7\text{V}$

Voltage across P can be given as
$V_P=I_{rms}\times Z_P$ .....(i)
For Box P, Impedence is given by
$Z_P=\sqrt{R_1^2+(X_C{)}^2}$ .......(ii)
Now if the maximum current flows through P and Q, then circuit is at resonance. Also at resonance $\omega =\frac{1}{\sqrt{LC}}$
$\omega =\frac{1}{\sqrt{4.9\times {10}^{-3}\times {10}^{-6}}}\omega =\frac{{10}^5}{7}$
From here we can find the value of $X_C$
$X_C=\frac{1}{\omega C}=\frac{1\times 7}{{10}^5\times {10}^{-6}}=70\Omega$
Putting this and value of $R_1$ in eq(ii)
$Z_P=\sqrt{{32}^2+{70}^2}=77\Omega$ ....(iii)
To calulate $I_{rms}$ we need to understand that the circuit is at resonance and it behaves as purely resistive ciruit.
So
$I_{rms}=\frac{V_{rms}}{R_1+R_2}=\frac{10}{68+32}=0.1A$ .....(iv)
From (iii) and (iv)
$V_P=0.1\times 77=7.7\text{V}$