The network shown in figure is a part of the circuit. (The battery has negligible resistance) At a certain instant the current I = 5A and is decreasing at a rate of 10^3 As^(-1). What is the potential difference between point B and A ?

The network shown in figure is a part of the circuit. (The battery has

1.	The network shown in figure is a part of the circuit. (The battery has negligible resistance) At a certain instant the current I = 5A and is decreasing at a rate of 10^3 As^(-1). What is the potential difference between point B and A ?
Answer» 5V 10V 15V 0V Solution : Taking direction from A to B , the kirchoff.s LAW , `V_A-IR+EPSILON-(-L(DI)/(dt))=V_B` `therefore -IR+epsilon+(LdI)/(dt)=V_B-V_A` Here `R=1 OMEGA , I=5A , (dI)/(dt) =10^3` A/s , L=5 mH, `epsilon`=15 V `therefore -5xx1+15+5xx10^(-3)xx10^3=V_B-V_A` `therefore -5+15+5=V_B-V_A` `therefore V_B-V_A` =15 V

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The network shown in figure is a part of the circuit. (The battery has negligible resistance) At a certain instant the current I = 5A and is decreasing at a rate of 10^3 As^(-1). What is the potential difference between point B and A ?

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