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Questions  

Consider the circuit diagram shown.

Column - IColumn - II
(i)Potential difference across A and D in steady state isp.2 V
(ii)Potential difference across capacitor in steady state isq.1.8 V
(iii)Value of Q for which no energy is stored across capacitor isr.0.2 V
(iv)Potential difference across A and B in steady state iss.143Ω

Now, match the given columns and select the correct option from the codes given below

a
i-q, ii-r, iii-s, iv-p
b
i-r, ii-p, iii-q, iv-s
c
i-q, ii-q, iii-s, iv-r
d
i-p, ii-p, iii-q, iv-s

detailed solution

Correct option is A

As 3  Ω and 7  Ω are in series across 6 v and in series potential divides in proportion to resistance.So VA −VD=R × XR +S = 3 ×  63+7  =1810  =  1.8  VHence (i) → (q)Similarly, VA −VB=P × VP +Q = 2 ×  62+4  =  2VVA −VB−VA −VD= VD −VB=2−1.8=0.2  VHence (ii) → (r)Energy stored in the capacitor will be zero if VB = VDi.e., PQ  =  RS  ⇒  2Q =37 or  Q=143 ΩSo (iii) → (s)Hence, [i-q], [ii-r], [iii-s] and [iv-p]

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