Question
Download Solution PDFIn the circuit shown above, the current through RL is
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFApplying Millman's theorem we get
\({V_{TH}} = \frac{{\frac{{{V_1}}}{{{R_1}}} + \frac{{{V_2}}}{{{R_2}}}}}{{\frac{1}{{{R_1}}} + \frac{1}{{{R_2}}}}}\)
\( = \frac{{\frac{{420}}{{120}} + \frac{{420}}{{60}}}}{{\frac{1}{{120}} + \frac{1}{{60}}}}\)
\( = \frac{{\frac{{840}}{{120}}}}{{\frac{3}{{120}}}}\)
= 420 V
\({R_{th}} = \frac{1}{{\left( {\frac{1}{{120}} + \frac{1}{{60}}} \right)}} = 40{\rm{\Omega }}\)
\({I_L} = \frac{{420}}{{70}}\)
= 6A
Last updated on Jul 2, 2025
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