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Five resistor each having value of 4 Omega are connected with ideal battery and ammeter as shown in figure. Find reading of ammeter. |
Answer» Solution :When 4 resistances each of VALUE`4 Omega` are connected in parallel equivalent resistance would be `(R)/(N) = (4)/(4) = 1 Omega`. Hence , above circuit can be redrawn as FOLLOWS :  Currne in the ammneter in above CLOSED loop is , `I= (epsilon )/(R_(eq))= (2)/(4 + 1 ) = (2)/(5) = 0.4 ` A  Equivalent resictance of pair of resistors,` = (4 xx 4)/(4 + 4 ) =(16)/(8) =2 Omega` Equivalent resistance of circuit , `R = (2 xx 2 )/(2 + 2 ) + 4 ` =1 + 4 = 5 `Omega` `therefore ` Current in AMMETER I ` = (epsilon)/(R)` `I =(2)/(3)` I = 0.4 A |
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