Question
Download Solution PDFFind the current flowing through the resistor R4?
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFApplying KCL at node A.
\(\frac{{V + {V_2}}}{5} + \frac{{V - 10}}{2} + \frac{V}{8} = 0 \)
\({V_2} = 3\;{V_m} = 3\frac{{\left( {V - 10} \right)}}{2} \times 2 \)
V2 = 3V – 30
\(\frac{{4V - 30}}{5} + \frac{{V - 10}}{2} + \frac{V}{8}= 0\)
32 V – 240 + 20 V – 200 + 5 V = 0
57 V = 440
\(V = \frac{{440}}{{57}}\)
\({\left. I \right|_{{R_3}}} = \frac{V}{8} = \frac{440}{57\times 8}=0.96~A\)
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