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## 3 resistors 4 ohm,6 ohm and 12 ohm are connected in parallel.The combination of the above resistors is connected in series to a resistance of 2 ohm and then to a battery of 6 volt. draw a circuit diagram and calculate: a)current in the main circuit. b)current flowing through each of the resistors in parallel. c)potential difference across the 2 ohm resistor. (diagram is not necessary).

5 years ago

Given-

R_{1}= 4ohm

R_{2 }= 6ohm

R_{3}= 12ohm

R_{s}= 2ohm (Series resistor)

Potential difference,V = 6V

Let the resistance of all three resistors which are connected in parallel be R_{p}

Then,

1/R_{p }= 1/R_{1}+1/R_{2}+ 1/R_{3}

1/R_{p }= 1/4 + 1/6 + 1/12

1/R_{p}= 6/12

or,R_{p}= 12/6 = 2ohm

Let total resistance be R_{t }= R_{p}+R_{s}

R_{t =}2 + 2 = 4ohm

a).So current,I in the main circuit will be V=IR

therefore, I = V/R = 6/4 = 1.5A which is total current.

b) Since in parallel the current will divided into three resistors connected in parallel.Let the current across all the three be I_{1},I_{2}and I_{3}.

Now,

I_{1 }= V/R_{1}(Since potential difference is same as they are connected in parallel)

I_{1 }= 6/4 = 1.5A

Similarly,I_{2}= V/R_{2}

I_{2}= 6/6 = 1A

and I_{3 }= 6/12 = 0.5A

c). V across 2ohm resistor will be-

V = I*R ( I will be total current)

V = 1.5 * 2 = 3V

5 years ago

Dear student i am again explaining the question you have asked.As per your question three resistors(4,6,12 ohm respectively)are connected in parallel and then connected in series with 2ohm resistor(R_{s}).Now,

a).The first part of question is asking about the current in the main circuit.So first we have to find the total resistance which we suppose to designated as R_{t}.For parallel combination

1/R_{p }= 1/R_{1}+ 1/R_{2}+ 1/R_{3 }

1/R_{p}= 1/4 + 1/6 + 1/12

R_{p}= 12/6 = 2ohm

For total resistance,Rt = R_{p}+ R_{s}(Since all three connected in series with the 2ohm resistor)

R_{t}= 2 + 2 = 4ohm

According to Ohm’s law,

V = I*R

I = V/R = 6/4 = 1.5A (Since the whole combination is connected to a battery of 6V)

b).Second part is asking the current through each resistor connected in parallel.Since in parallel combination current will get divided into three part each through the three resistors(I_{1},I_{2}and I_{3}).Voltage in such combination remains same.

So I_{1}= V/R_{1}, I_{2}= V/R_{2}and I_{3 }= V/R_{3}which we calculated as explained above.

3).Third part is asking the voltage across 2ohm resistor.Let it be V_{s}

V_{s}= I * R_{s }

For that main current which is calculated as above will be used as this resistor is connected in series so the main current will passes through it also.

V_{s}= 1.5 * 2 = 3V

5 years ago

lets find parallel resistance first= 24/12 = 2ohmp = 1/4 + 1/6 + 1/12=12/24Rp1/R3 + 1/R2 + 1/R1= 1/Rp 1/Rtotal resistance=Rp + Rs = 2 + 2 = 4ohmcurrent in circuit=i=V/R=6/4=1.5ampcurrent flow is diff in all parallel resistors and potential diff is same in all parallel resistors but not same as series resistor.V_{p}potential diff across parallel resistors,V_{s}potential diff aross series resistorV=V_{p}+V_{s}V_{s}=I(2)=(1.5)2=3VV_{p}=V-V_{s}=6-3=3Vcurrent through R_{1 }is I_{1}=V_{p}/R_{1}=3/4 ampcurrent through R_{2 }is I_{2}=V_{p}/R_{2}=3/6=1/2ampcurrent through R_{3 }is I_{3}=V_{p}/R_{3}=3/12=1/4ampapprove if it is r8

5 years ago

Dear student,Please find the attached solution to your problem.The resultant resistance from the parallel arrangement1/R_{p}= 1/4+ 1/6 + 1/12 = 6/12Hence, R_{p}= 2ΩNow total resistance in circuit = R_{p}+ R_{s}= 2 + 2 = 4Ωa). current in circuit, I = V/R = 6/4 = 1.5 Ab). Now the potential difference across the circuit will be the sum of potential drop across the two arrangements.Hence, V = V_{p}+ V_{s}V_{s}= I(2) = (1.5)2 = 3VV_{p}= V – V_{s}= 6 – 3 = 3Vcurrent through R_{1}, I_{1}= V_{p}/R_{1}= 0.75 Acurrent through R_{2}, I_{2}= V_{p}/R_{2}= 3/6 = 0.5 Acurrent through R_{3}, I_{3}= V_{p}/R_{3}= 3/12 = 0.25 Ac). We have already calculated, V_{s}= 3 VThanks and regards,Kushagra

5 months ago

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