This question was previously asked in

ESE Electronics 2014 Paper 2: Official Paper

Option 1 : 2.39 V and 0.2 Watt

CT 3: Building Materials

2894

10 Questions
20 Marks
12 Mins

__Concept:__

**RMS value**

It is the steady equivalent value of a time-varying waveform which could also develop the same amount of heat given by the original waveform for a definite period in the circuit.

It is defined as:

\({V_{RMS}} = \sqrt {\frac{1}{T}\mathop \smallint \nolimits_0^T {{\left[ {V\left( t \right)} \right]}^2}dt} \)

RMS value for a full-wave is given by

\({V_{RMS}} = \frac{{{V_m}}}{{\sqrt 2 }}\)

The average value is given by

\({V_{avg}} = \frac{{2{V_m}}}{\pi }\)

V: voltage of the waveform is considered

**NOTE: **If the current is taken then also the same formulae applied.

__Calculation:__

Given that the current supplied by the center-tapped transformer is 100 mA. i.e, I_{DC} = 100 mA

R_{L }= 20 Ω

The secondary resistance of the transformer is 1 Ω

The circuit is shown as

**Power calculation**

Power is given by

\(P = VI = \frac{{{V^2}}}{R} = {I^2}R\)

Given load current is 100 mA

P = (100 × 10^{-3})^{2} × 20

P = 10^{4} × 10^{-6} × 20

P = 10^{-1} × 2

**P = 0.2 W**

Following the option elimination method, we get the answer as option 1

**Calculation of the RMS voltages across each half secondary**

\({I_{DC}} = \frac{{2{I_m}}}{\pi } = 100mA\)

\({I_m} = \frac{\pi }{2} \times 100mA\)

\({V_0}_{peak} = {I_m}\left( {20} \right)\)

\(= \frac{\pi }{2} \times 100mA \times 20\)

\({V_0}_{peak} = \pi \)

Applying the KVL

- V_{m1} + (0.5I_{m}) + (0.5I_{m}) + π = 0

V_{m1} = I_{m }+ π

V_{m1} = 0.157 + 3.14

**V _{m1} = 3.297 V**

RMS value is

\({V_{m1\left( {RMS} \right)}} = \frac{{{V_{m1}}}}{{\sqrt 2 }}\)

\({V_{m1\left( {RMS} \right)}} = \frac{{3.297}}{{\sqrt 2 }}\)

**V _{m1(RMS)} = 2.336 V**

From the given options it is near to option 1.