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Find the value of R in Fig. so that there is no current in the 15Omega resistor. |
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Answer» Solution :This is the Wheatstone bridge with the GALVANOMETER REPLACED by `15Omega` resistor. There bridge is balanced because there is no CURRENT in `15Omega` resistor, hence, `20//10=40R""THEREFORE R=(40xx10Omega)/(20)=20Omega`. |
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