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MCQEasyJEE 2024Diodes & Rectifiers

JEE Physics 2024 Question with Solution

In the given circuit, the voltage across load resistance RLR_L is:

  • A

    8.75 V8.75 \, \text{V}

  • B

    9.00 V9.00 \, \text{V}

  • C

    13.50 V13.50 \, \text{V}

  • D

    14.00 V14.00 \, \text{V}

Answer

Correct answer:A

Step-by-step solution

Standard Method

Given: RD1=1.5 kΩR_{D1} = 1.5 \, \text{k}\Omega, RL=2.5 kΩR_L = 2.5 \, \text{k}\Omega, source voltage V=15 VV = 15 \, \text{V}, diode drops VD1=0.3 VV_{D1} = 0.3 \, \text{V} and VD2=0.7 VV_{D2} = 0.7 \, \text{V}.

Find: Voltage across load resistance RLR_L.

Both diodes are forward biased, so their total voltage drop is subtracted from the source voltage.

Vresistive=V−(VD1+VD2)V_{\text{resistive}} = V - (V_{D1} + V_{D2})Vresistive=15 V−(0.3 V+0.7 V)=14 VV_{\text{resistive}} = 15 \, \text{V} - (0.3 \, \text{V} + 0.7 \, \text{V}) = 14 \, \text{V}

The total resistance in the resistive part is

Rtotal=RD1+RL=1.5 kΩ+2.5 kΩ=4.0 kΩR_{\text{total}} = R_{D1} + R_L = 1.5 \, \text{k}\Omega + 2.5 \, \text{k}\Omega = 4.0 \, \text{k}\Omega

Using Ohm's law, the circuit current is

I=VresistiveRtotal=14 V4.0 kΩ=144000 A=0.0035 A=3.5 mAI = \frac{V_{\text{resistive}}}{R_{\text{total}}} = \frac{14 \, \text{V}}{4.0 \, \text{k}\Omega} = \frac{14}{4000} \, \text{A} = 0.0035 \, \text{A} = 3.5 \, \text{mA}

Now the voltage across the load resistance is

VL=I×RLV_L = I \times R_LVL=3.5 mA×2.5 kΩ=0.0035 A×2500 Ω=8.75 VV_L = 3.5 \, \text{mA} \times 2.5 \, \text{k}\Omega = 0.0035 \, \text{A} \times 2500 \, \Omega = 8.75 \, \text{V}

Therefore, the voltage across the load resistance is 8.75 V8.75 \, \text{V}. The correct option is A.

KVL-Based Detailed Solution

Given: A series circuit with a 15 V15 \, \text{V} source, a germanium diode with drop 0.3 V0.3 \, \text{V}, a silicon diode with drop 0.7 V0.7 \, \text{V}, a resistor 1.5 kΩ1.5 \, \text{k}\Omega, and load resistance RL=2.5 kΩR_L = 2.5 \, \text{k}\Omega.

Find: Voltage across RLR_L.

Apply Kirchhoff's voltage law to the loop:

Vs=VGe+VSi+IRS+IRLV_s = V_{Ge} + V_{Si} + I R_S + I R_LVs=VGe+VSi+I(RS+RL)V_s = V_{Ge} + V_{Si} + I(R_S + R_L)

First calculate the total diode drop:

Vdiodes=0.3 V+0.7 V=1.0 VV_{\text{diodes}} = 0.3 \, \text{V} + 0.7 \, \text{V} = 1.0 \, \text{V}

So the voltage across the resistors is

Vresistors=15 V−1.0 V=14 VV_{\text{resistors}} = 15 \, \text{V} - 1.0 \, \text{V} = 14 \, \text{V}

Now calculate total resistance:

Rtotal=1.5 kΩ+2.5 kΩ=4.0 kΩ=4000 ΩR_{\text{total}} = 1.5 \, \text{k}\Omega + 2.5 \, \text{k}\Omega = 4.0 \, \text{k}\Omega = 4000 \, \Omega

Hence current in the series circuit is

I=144000 A=3.5×10−3 AI = \frac{14}{4000} \, \text{A} = 3.5 \times 10^{-3} \, \text{A}

Voltage across the load resistance becomes

VRL=I×RLV_{R_L} = I \times R_LVRL=(3.5×10−3 A)(2.5×103 Ω)V_{R_L} = (3.5 \times 10^{-3} \, \text{A})(2.5 \times 10^3 \, \Omega)VRL=8.75 VV_{R_L} = 8.75 \, \text{V}

Therefore, the voltage across the load resistance is 8.75 V8.75 \, \text{V}.

Common mistakes

  • Subtracting only one diode drop from the source voltage is incorrect because both forward-biased diodes contribute to the total drop. Add 0.3 V0.3 \, \text{V} and 0.7 V0.7 \, \text{V} before finding the resistor voltage.

  • Using only RLR_L to calculate the circuit current is wrong because the current flows through the entire series resistance. First use RD1+RLR_{D1} + R_L to find the current, then use that current to find the voltage across RLR_L.

  • Mixing kΩ\text{k}\Omega and Ω\Omega carelessly can produce an incorrect current. Convert consistently, for example 4.0 kΩ=4000 Ω4.0 \, \text{k}\Omega = 4000 \, \Omega, before applying Ohm's law.

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