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ln the given circuit, calculate : (i) the total resistance of the circuit (ii) the current through the circuit, and (iii) the potential difference across R_(1) and R_(2) |
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Answer» Solution :Given, the resistance of conductor `R_(1) = 4Omega` The resistance of conductor `R_(2) = 20Omega` (i) The total resistance of the circuit R= `R_(1) + R_(2)`(series COMBINATION) (i) Now by Ohm.s Law The current through the circuit `V=IRI= (V)/(R) = (6)/(24) = (1)/(4) =0.25 ` Amp. (iii) The potential difference across the TWO TERMINALS of the battery = 6 V . `V_(1)= IxxR_(1)` = `0.25xx4= 1V` ( Potential difference across the conductor `R_(1))` `V_(2) =IxxR_(2)` = `0.25xx20= 5V` ( Potential difference across the conductor `R_(2))` |
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