The effective capacitance between the points A and B in the circuit shown in (μ F) is (capacitance of each capacitor is(μ F)

Question

The effective capacitance between the points A and B in the circuit shown in (μ F) is (capacitance of each capacitor is(μ F) (A) 2 (B) 4 (C) 3 (D) 0.4

Tardigrade Admin · Accepted Answer

Capacitance of 2 capacitors in parallel = 1 + 1 = 2μF Capacitance of 1 μF, 2μF and 1 μF in series is given by, (1/C) = (1/1) + (1/2) + (1/1) = (2 + 1 + 2/2) = (5/2) i.e. C = (2/5) = 0.4 μ F

Q. The effective capacitance between the points $A$ and $B$ in the circuit shown in $(μ F)$ is (capacitance of each capacitor is $(μ F)$

2

4

3

0.4

Capacitance of 2 capacitors in parallel = 1 + 1 = 2 $μ$ F
Capacitance of 1 $μ$ F, 2 $μ$ F and 1 $μ$ F in series is given by, $\frac{1}{C} = \frac{1}{1} + \frac{1}{2} + \frac{1}{1} = \frac{2 + 1 + 2}{2} = \frac{5}{2}$
$i . e .$ C = $\frac{2}{5} = 0.4 μ F$

Q. The effective capacitance between the points A and B in the circuit shown in (μF) is (capacitance of each capacitor is(μF)

2

4

3

0.4

Capacitance of 2 capacitors in parallel = 1 + 1 = 2μF Capacitance of 1 μF, 2μF and 1 μF in series is given by, C1​=11​+21​+11​=22+1+2​=25​ i.e. C = 52​=0.4μF

Q. The effective capacitance between the points $A$ and $B$ in the circuit shown in $(μ F)$ is (capacitance of each capacitor is $(μ F)$

Capacitance of 2 capacitors in parallel = 1 + 1 = 2 $μ$ F
Capacitance of 1 $μ$ F, 2 $μ$ F and 1 $μ$ F in series is given by, $\frac{1}{C} = \frac{1}{1} + \frac{1}{2} + \frac{1}{1} = \frac{2 + 1 + 2}{2} = \frac{5}{2}$
$i . e .$ C = $\frac{2}{5} = 0.4 μ F$