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The current pasing through the ideal ammeter in the circuit given below is A. 1.25 AB. `1A`C. 0.75 AD. 0.5 A |
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Answer» Correct Answer - D Here `2Omega` and `2 Omega` are in parallel. `(1)/(R )=(1)/(2)+(1)/(2) rArr R =(2xx2)/(2+2)=1 Omega` Now, internal resistances `1Omega, 2Omega`and resistance R are in series. `therefore " " R_("net")=1Omega +2 Omega+4Omega+1Omega = 8 Omega` Hence, current `l=(V)/(R )=(4)/(8)=0.5 A` |
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