Question 1:The equivalent resistance of the given circuit is calculated byReq = 20 k || 20 k Req =( 20 k )( 20 k )20 k + 20 k Req = 10 k The equivalence capacitance of the given circuit is calculated byCeq = (1 F + 1 F ) || 2 F Ceq = 2 F || 2 F Ceq =( 2 F )( 2 F )2 F + 2 F Ceq = 1 FHence, it turns out that for t 0 , we have RC circuit connected in series with R = 10 k andC = 1 F , as shown in the below figureUsing Kirchoffs voltage law (KVL), we havevs Ri ( t ) vC ( t ) = 0Solving the above equation gives the voltage across the capacitor asvc ( t ) = vs (1 e t / RC )t(10103 )(1106 F ) vc ( t ) = (10 V ) 1 e vc ( t ) = 10 (1 e 100t ) VHence, for t 0 , the required voltage is calculated to bev ( t ) = vs vc ( t ) v ( t ) = 10 V 10 (1 e100t ) V v ( t ) = 10e100t VQuestion 2:The equivalent resistance is calculated byReq = ( 20 k || 20 k ) + 10 k Req =( 20 k )( 20 k ) + 10 k 20 k + 20 k ( 20 )( 20 ) + 10 Req =20 + 20 Req = 20 k The equivalent inductance is calculated byLeq = ( 2 mH || 2 mH ) + 1 mH Leq =( 2 mH )( 2 mH ) + 1 mH2 mH + 2 mH( 2 )( 2 ) + 1 Leq =2+2 Leq = 2 mHHence, it turns out that for t 0 , we have RL circuit connected in series with R = 20 k ...
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