When swithch S is off, both resistances are included in the circuit.
The resistance of the bulb RB = V2/P = 1002/500 = 20\Omega
When connected across a 200V supply with a resistor R in series, the power drawn remains the same. If V1 be the p.d across the bulb,V12 = P {RB} = 500 * 20 = 104;
V1 = 100V, . Therefore the p.d across R =200-100=100 V. Therefore R=RB=20Ω
The capacitance of a capacitor is a constant.
Also, V q
The discharging equation for capacitor is
A similar equation can be written for potential difference.
With a known resistance in one of the gaps, the meter bridge is used to determine the value of an unknown resistance by the formula,
where l is the null point.
For the given circuit , with cell e m f = 10 V and internal resistance = , which of the following is correct?
The circuit is reduced to find equivalent resistance as follows;
In (1), 2Ω, 4 Ω, and 2 Ω are in series. Their equivalent resistance is 8 Ω. In (2), the two 8 Ω resistors are in parallel and their equivalent resistance is 4 Ω. In (3) 2 Ω, 4 Ω, and 2 Ω are in series. Their equivalent resistance is 8 Ω which is in parallel with the 8 Ω resistance as shown in (4). The total resistance in the circuit
R= 3+4+2+1( internal resistance)=10 Ω. The current through the 3 Ω resistor
I = E/R = 10/10
I = 1A
V3 and combination of V1 and V2 (V1 + V2) are connected in parallel so potential in V3 must be equal to potential in (V1 + V2)
the potential difference across AB = = +.