Question

Find current through the battery in the circuit shown in fig :
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Hint:

When you see diodes in a DC circuit, the very first step is to check their biasing. A forward-biased ideal diode is a short circuit (wire), and a reverse-biased ideal diode is an open circuit (a break in the wire). This simplification is the key to solving such problems.

Answer 1

Step 1: Understanding the Concept:
This problem requires analyzing a circuit containing diodes and resistors in parallel. The key is to first determine the biasing of each diode (forward or reverse) to see which branches will conduct current. Then, the circuit can be simplified to calculate the total current using Ohm's law.

Step 2: Detailed Explanation:
1. Analyze the Diode Biasing:
The circuit has a 10 V battery. The positive terminal of the battery is connected to the p-side (anode, the triangle part) of all three diodes. The negative terminal is connected (through the resistors) to the n-side (cathode, the line part) of all three diodes.
Since the p-side of each diode is at a higher potential than its n-side, all three diodes are forward-biased.
2. Simplify the Circuit:
Assuming the diodes are ideal, a forward-biased diode acts as a closed switch, meaning it behaves like a wire with zero resistance. Therefore, we can replace all three diodes with connecting wires.
After this simplification, the circuit consists of three resistors connected in parallel across the 10 V battery: \begin{itemize} \item \(R_1 = 6 \, \Omega\)
\item \(R_2 = 2 \, \Omega\)
\item \(R_3 = 3 \, \Omega\)
\end{itemize} 3. Calculate the Equivalent Resistance (\(R_{eq}\)):
For resistors in parallel, the reciprocal of the equivalent resistance is the sum of the reciprocals of the individual resistances: \[ \frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} \] \[ \frac{1}{R_{eq}} = \frac{1}{6} + \frac{1}{2} + \frac{1}{3} \] To add these fractions, we find a common denominator, which is 6: \[ \frac{1}{R_{eq}} = \frac{1}{6} + \frac{3}{6} + \frac{2}{6} = \frac{1+3+2}{6} = \frac{6}{6} = 1 \, \Omega^{-1} \] Therefore, the equivalent resistance is: \[ R_{eq} = 1 \, \Omega \] 4. Calculate the Total Current:
The current through the battery is the total current flowing through the circuit. Using Ohm's Law, \(V = I_{total} \cdot R_{eq}\): \[ I_{total} = \frac{V}{R_{eq}} \] \[ I_{total} = \frac{10 \, \text{V}}{1 \, \Omega} = 10 \, \text{A} \]

Step 3: Final Answer:
The current through the battery is 10 A.