find the equivalent resistance across the two ends a and b of the following circuit
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Answered by
6
Hey Mate
your answer is ---
equivalent resistance is --
1/R = 1/5+1/10+1/5
=> 1/R = 2+1+2/10
=> 1/R = 5/10
=> R = 10/5
=> R = 2ohm
hence, equivalent resistance is 2ohm
hope it help you
your answer is ---
equivalent resistance is --
1/R = 1/5+1/10+1/5
=> 1/R = 2+1+2/10
=> 1/R = 5/10
=> R = 10/5
=> R = 2ohm
hence, equivalent resistance is 2ohm
hope it help you
Anonymous:
:D
Answered by
5
Yeah sure for your help
********************
®®®®®®®®®®®®®®®
Here is your answer which you are searching for
✍✍
♦ In between of the circuit you see 5 Omega and 5 Omega are connected in series
♦ In series we use the formula Rs = R1+R2 ......Rn
♦ Resistance in series = 5 + 5 = 10 Omega
♦ Now you see Resistance are connected in parallel means that it have connected with two ends of the circuit
♦ Formula for Resistance in parallel is 1/R1 +1/R2...........1/R/n
♦ 1/R = 1/5+ 1/10+ 1/5
♦ When we calculate this our answer becomes 2omega
♦ 2 Omega is the answer of your question
*********************
®®®®®®®®®®®®®®®®®
✍✍
Be Brainly
Be Proud
Together we can change the world ❤
<< Warm Regards
@ Brainlestuser
********************
®®®®®®®®®®®®®®®
Here is your answer which you are searching for
✍✍
♦ In between of the circuit you see 5 Omega and 5 Omega are connected in series
♦ In series we use the formula Rs = R1+R2 ......Rn
♦ Resistance in series = 5 + 5 = 10 Omega
♦ Now you see Resistance are connected in parallel means that it have connected with two ends of the circuit
♦ Formula for Resistance in parallel is 1/R1 +1/R2...........1/R/n
♦ 1/R = 1/5+ 1/10+ 1/5
♦ When we calculate this our answer becomes 2omega
♦ 2 Omega is the answer of your question
*********************
®®®®®®®®®®®®®®®®®
✍✍
Be Brainly
Be Proud
Together we can change the world ❤
<< Warm Regards
@ Brainlestuser
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