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).

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Aarti Gupta · Answer

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

runo · Answer

will someone post the proper answer.plzz

Aarti Gupta · Answer

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/R₁ + 1/R₂ + 1/R₃
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(I₁,I₂ and I₃).Voltage in such combination remains same.
So I₁ = V/R₁ , I₂ = V/R₂ and I₃= V/R₃ 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

noogler · Answer

lets find parallel resistance first = 24/12 = 2ohm p = 1/4 + 1/6 + 1/12=12/24R p 1/R 3 + 1/R 2 + 1/R 1 = 1/R p 1/R total resistance= R p + R s = 2 + 2 = 4ohm current in circuit=i=V/R=6/4=1.5amp current 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 resistor V=V p +V s V s =I(2)=(1.5)2=3V V p =V-V s =6-3=3V current through R 1 is I 1 =V p /R 1 =3/4 amp current through R 2 is I 2 =V p /R 2 =3/6=1/2amp current through R 3 is I 3 =V p /R 3 =3/12=1/4amp approve if it is r8

Kushagra Madhukar · Answer

Dear student, Please find the attached solution to your problem. The resultant resistance from the parallel arrangement 1/R p = 1/4+ 1/6 + 1/12 = 6/12 Hence, R p = 2Ω Now total resistance in circuit = R p + R s = 2 + 2 = 4Ω a). current in circuit, I = V/R = 6/4 = 1.5 A b). 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 = 3V V p = V – V s = 6 – 3 = 3V current through R 1 , I 1 = V p /R 1 = 0.75 A current through R 2 , I 2 = V p /R 2 = 3/6 = 0.5 A current through R 3 , I 3 = V p /R 3 = 3/12 = 0.25 A c). We have already calculated, V s = 3 V Thanks and regards, Kushagra

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