Question

two capacitors of 2 μf and 4 μf are connected in parallel. a third capacitor of 6 μf is connected in series. the combination is connected across a 12 v battery. the voltage across 2 μf capacitor is (1)2 v (2)8 v (3)6 v (4)1 v

Explanation:

Step 1: Find equivalent capacitance of parallel capacitors

First, the 2 μF and 4 μF capacitors are in parallel. The formula for equivalent capacitance (\(C_{eq1}\)) of parallel capacitors is \(C_{eq1}=C_1 + C_2\). So, \(C_{eq1}=2\ \mu\text{F}+4\ \mu\text{F}=6\ \mu\text{F}\).

Step 2: Find equivalent capacitance of series combination

Now, this \(6\ \mu\text{F}\) (equivalent of parallel) is in series with the 6 μF capacitor. The formula for equivalent capacitance (\(C_{eq}\)) of series capacitors is \(\frac{1}{C_{eq}}=\frac{1}{C_{eq1}}+\frac{1}{C_3}\). Substituting \(C_{eq1} = 6\ \mu\text{F}\) and \(C_3=6\ \mu\text{F}\), we get \(\frac{1}{C_{eq}}=\frac{1}{6}+\frac{1}{6}=\frac{2}{6}=\frac{1}{3}\), so \(C_{eq}=3\ \mu\text{F}\).

Step 3: Find total charge in the circuit

Using \(Q = C_{eq}V\) (where \(V = 12\ \text{V}\) is the battery voltage), \(Q=3\ \mu\text{F}\times12\ \text{V}=36\ \mu\text{C}\). In series, the charge on each capacitor (the 6 μF equivalent of parallel and the 6 μF capacitor) is the same as the total charge. So the charge on the 6 μF equivalent (which is the parallel combination of 2 μF and 4 μF) is \(36\ \mu\text{C}\).

Step 4: Find voltage across 2 μF capacitor

For the parallel combination, the voltage across each capacitor (2 μF and 4 μF) is the same as the voltage across their equivalent (\(C_{eq1}\)). Using \(V=\frac{Q}{C}\), for the parallel equivalent (\(C_{eq1}=6\ \mu\text{F}\)) with charge \(Q = 36\ \mu\text{C}\), the voltage \(V_{eq1}=\frac{36\ \mu\text{C}}{6\ \mu\text{F}} = 6\ \text{V}\). Since the 2 μF capacitor is in parallel with the 4 μF capacitor, the voltage across the 2 μF capacitor is also 6 V? Wait, no, wait. Wait, the parallel combination (2 and 4 μF) is in series with 6 μF. So the charge on the parallel combination is equal to the charge on the 6 μF capacitor. Wait, let's re - check.

Wait, when two capacitors are in series, the charge on each is the same. The parallel combination (2 and 4 μF) has an equivalent capacitance of 6 μF. Then this 6 μF is in series with another 6 μF. So the total capacitance is \(\frac{6\times6}{6 + 6}=3\ \mu\text{F}\), total charge \(Q = C_{total}\times V=3\times12 = 36\ \mu\text{C}\). So the charge on the 6 μF (parallel equivalent) is 36 μC, and the charge on the other 6 μF is also 36 μC. Now, for the parallel combination (2 and 4 μF), the voltage across them is \(V=\frac{Q}{C_{eq1}}=\frac{36\ \mu\text{C}}{6\ \mu\text{F}} = 6\ \text{V}\). Since in parallel, the voltage across each capacitor (2 μF and 4 μF) is the same as the voltage across the parallel combination. So the voltage across the 2 μF capacitor is 6 V? Wait, but let's think again. Wait, the 2 μF and 4 μF are in parallel, so their voltages are equal, and the charge on 2 μF is \(Q_1 = C_1V\), charge on 4 μF is \(Q_2 = C_2V\), and \(Q_1+Q_2=Q_{eq1}=36\ \mu\text{C}\). Let \(V\) be the voltage across them. Then \(2V + 4V=36\), \(6V = 36\), \(V = 6\ \text{V}\). So the voltage across the 2 μF capacitor is 6 V.

Answer:

(3) 6 V