Three ideal diodes and resistors connected to the cell of negligible internal resistance is as shown. Find the current passing through the \( 10\Omega \) resistor.
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For ideal diode circuits, first identify which diodes are forward biased and reverse biased. Forward biased diodes act as short circuits, while reverse biased diodes act as open circuits.
Step 1: Identify the polarity of the battery.
The cell of \( 10V \) makes the left side of the circuit at higher potential than the right side. Therefore, current tends to flow from left to right through conducting branches. Step 2: Check diode \( D_1 \).
Diode \( D_1 \) is connected in such a way that it is forward biased for current from left to right. Hence, \( D_1 \) conducts and acts like a short circuit. Step 3: Check diode \( D_2 \).
Diode \( D_2 \) is oppositely connected, so it is reverse biased for the given battery polarity. Hence, \( D_2 \) does not conduct and the \( 40\Omega \) branch is open. Step 4: Check diode \( D_3 \).
Diode \( D_3 \) is also forward biased for current from left to right. Hence, \( D_3 \) conducts and the branch containing \( 20\Omega \) resistor becomes active. Step 5: Find the equivalent resistance of active parallel branches.
The conducting branches contain two \( 20\Omega \) resistors in parallel:
\[
R_p = \frac{20 \times 20}{20 + 20}
\]
\[
R_p = 10\Omega
\] Step 6: Add the series \( 10\Omega \) resistor.
This equivalent resistance is in series with the \( 10\Omega \) resistor connected with the battery. Therefore, total resistance is:
\[
R_{\text{total}} = 10 + 10 = 20\Omega
\] Step 7: Calculate the current through the \( 10\Omega \) resistor.
Since the \( 10\Omega \) resistor is in series with the battery, the current through it is:
\[
I = \frac{V}{R_{\text{total}}}
\]
\[
I = \frac{10}{20} = 0.5A
\]
Hence, the current passing through the \( 10\Omega \) resistor is \( 0.5A \).
