In the given circuit, the current (I) through the battery will be
- 1.5A
- 2.5A
- 1A
- 2A
The Correct Option is A
Solution and Explanation
Understanding the Problem
In the given figure, the diodes \( D_1 \) and \( D_3 \) are forward biased and the diode \( D_2 \) is reversed biased. We need to find the current through the battery.
Solution
1. Diode Analysis:
The diodes \( D_1 \) and \( D_3 \) are forward biased, so they act as short circuits (negligible resistance). The diode \( D_2 \) is reverse biased, so it acts as an open circuit (infinite resistance).
2. Equivalent Circuit:
The circuit simplifies to two resistors in parallel: one \( 20 \, \Omega \) resistor and one \( 10 \, \Omega \) resistor. The reverse-biased \( D_2 \) branch is effectively removed from the circuit.
3. Equivalent Resistance (\(R_{eq}\)):
The equivalent resistance of parallel resistors is given by:
\( R_{eq} = \frac{R_1 R_2}{R_1 + R_2} \)
Substituting the values:
\( R_{eq} = \frac{(20)(10)}{(20) + (10)} = \frac{200}{30} = \frac{20}{3} \, \Omega \)
4. Current Through the Battery (I):
Using Ohm's law, \( I = \frac{V}{R} \), where \( V = 10 \, \text{V} \):
\( I = \frac{10}{\frac{20}{3}} = 10 \times \frac{3}{20} = \frac{30}{20} = 1.5 \, \text{A} \)
Final Answer
The current through the battery is \( 1.5 \, \text{A} \).
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