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A 10 V battery of negilgible internal resistance is connected across a 200V battery and a resistance of `38 Oemga` as shown in figure. . Find the value of current in the circuit. |
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Answer» Here, `epsilon_(1)=200V, epsilon_(2)=10 V , R=38 Omega , I=?` `I=(epsilon_(1)-epsilon_(2))/R = (200-10)/(38)=5A` |
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