Please answer this question

### Asked by tanyajain164 | 22nd 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 R_{2} , R_{3} and R_{4} are shorted at B and C .
Further, ends of resitor R_{3} are dircectly connected to terminal points of battery at A and D.
Hence there will not be any current flow in resistors R_{2} and R_{4} . Hence they can be removed in the circuit .
Fig(2) shows the final circuit after removing R_{2} and R_{4} .
Equivalenet resistance of the circuit is series combination of R_{1} and R_{5} in parallel to R_{3} .
R_{eq} = ( R_{1} + R_{5} ) || R_{3} = [ (6 + 6 ) || 12 ] Ω = 6 Ω
-----------------------------------------------
(1) Equivalent resistance of circuit is 3 Ω
Above statement is false because , we have calculated the equivalent resistance as 6 Ω
------------------
(2) Current flow through R_{3} is 4 A
Above statement is false because as seen from fig(1) and fig(2) , resistor R_{3} = 12 Ω is connected directly to 12 V battery .
Hence current flowing through R_{3} is 1 A
-------------------------------------------------
(3) Voltage drop across R_{1} is 6 V
Above statement is true as explained below.
Resistors R_{1} and R_{5} are in series and directly connected to 12 V battery [ Refer fig(1) and fig (2) ]
Hence current flow through R_{1} and R_{5} = [ 12 V / ( 6 + 6 ) Ω ] = 1 A
Hence potential drop across R_{1} = ( 6 × 1 ) = 6 V
----------------------------------------------------
(4) Current flow through R_{2} = 1 A
Above statement is false , current does not flow through R_{2}

_{2}, R

_{3}and R

_{4}are shorted at B and C .

_{3}are dircectly connected to terminal points of battery at A and D.

_{2}and R

_{4}. Hence they can be removed in the circuit .

_{2}and R

_{4}.

_{1}and R

_{5}in parallel to R

_{3}.

_{eq}= ( R

_{1}+ R

_{5}) || R

_{3}= [ (6 + 6 ) || 12 ] Ω = 6 Ω

_{3}is 4 A

_{3}= 12 Ω is connected directly to 12 V battery .

_{3}is 1 A

_{1}is 6 V

_{1}and R

_{5}are in series and directly connected to 12 V battery [ Refer fig(1) and fig (2) ]

_{1}and R

_{5}= [ 12 V / ( 6 + 6 ) Ω ] = 1 A

_{1}= ( 6 × 1 ) = 6 V

_{2}= 1 A

_{2}

### Answered by Thiyagarajan K | 22nd Jan, 2022, 11:37: AM

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