Request a call back Asked by tanyajain164 | 22 Jan, 2022, 09:44: AM Expert Answer Given circuit is redrawn in fig.(1) for better understanding .
As can be seen from fig.(1) , resistors R2 , R3 and R4 are shorted at B and C .

Further, ends of resitor R3 are dircectly connected to terminal points of battery at A and D.

Hence there will not be any current flow in resistors R2 and R4 . Hence they can be removed in the circuit .

Fig(2) shows the final circuit after removing R2 and R4 .
Equivalenet resistance of the circuit is series combination of R1 and R5 in parallel to R3 .

Req = ( R1 + R5 ) || R3  = [ (6 + 6 ) || 12 ] Ω  = 6 Ω

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(1) Equivalent resistance of circuit is 3 Ω

Above statement is false because , we have calculated the equivalent resistance as 6 Ω

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(2) Current flow through R3 is 4 A

Above statement is false because as seen from fig(1) and fig(2) , resistor R3 = 12 Ω  is connected directly to 12 V battery .
Hence current flowing through R3 is 1 A

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(3) Voltage drop across R1 is 6 V

Above statement is true as explained below.

Resistors R1 and R5 are in series and directly connected to 12 V battery [ Refer fig(1) and fig (2) ]

Hence current flow through R1 and R5 = [ 12 V / ( 6 + 6 ) Ω ] = 1 A

Hence potential drop across R1 = ( 6 × 1 ) = 6 V

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(4) Current flow through R2 = 1 A

Above statement is false , current does not flow through R2
Answered by Thiyagarajan K | 22 Jan, 2022, 11:37: AM
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