As per the given circuit, the value of current through the battery will be A. [Figure: a 10 V battery on the left (positive at top). From the top-left node three branches go right: D₁ (reverse biased) and 6 Ω to the top-right node; D₂ (forward biased) and 6 Ω to the top-right node; a diagonal of 10…

Question

As per the given circuit, the value of current through the battery will be A. [Figure: a 10 V battery on the left (positive at top). From the top-left node three branches go right: D₁ (reverse biased) and 6 Ω to the top-right node; D₂ (forward biased) and 6 Ω to the top-right node; a diagonal of 10 Ω, D₃ (forward biased) and 6 Ω down to the bottom-right node. A 10 Ω resistor joins the top-right node to the bottom-right node, and a 2 Ω resistor joins the bottom-right node back to the battery.]

Accepted Answer

1. D₁ blocks. D₂ and D₃ are ideal and conduct. Path via D₂: 6 Ω then 10 Ω down the right side: 16 Ω. Diagonal path via D₃: 10+6=16 Ω. Parallel: 8 Ω. Add the 2 Ω: 10 Ω. I=(10)/(10)=1 A

Step-by-step solution