Electronics - Parallel Circuits - Discussion
Discussion Forum : Parallel Circuits - General Questions (Q.No. 4)
4.
What is the total power loss if 2 k
and 1 k parallel-connected resistors have an IT of 3 mA?
and 1 k parallel-connected resistors have an IT of 3 mA?6 
W
W36 W
W6 mW
36 mW
Discussion:
39 comments Page 1 of 4.
Moksha said:
7 years ago
p = I^2R.
1/R=(1+2)/(1*2) = 3/2,
R = 2/3k ohms,
p = (3m)^2 * 2/3k = 6mw.
1/R=(1+2)/(1*2) = 3/2,
R = 2/3k ohms,
p = (3m)^2 * 2/3k = 6mw.
(1)
Sameer said:
7 years ago
Power=Voltage * Current.
Voltage= Current * Resistance,
Resistors in parallel (1/R=1/R1+1/R2) OR (R=R1*R2/R1+R2),
so, 1/R= 1/2+1/1 = 3/2KOhm , R= 2/3 KOhm,
P=V*I = (I*R)* I.
P= I*I*R,
P=3*3*2/3,
P=9*2/3,
P=18/3,
P=6mW.
Voltage= Current * Resistance,
Resistors in parallel (1/R=1/R1+1/R2) OR (R=R1*R2/R1+R2),
so, 1/R= 1/2+1/1 = 3/2KOhm , R= 2/3 KOhm,
P=V*I = (I*R)* I.
P= I*I*R,
P=3*3*2/3,
P=9*2/3,
P=18/3,
P=6mW.
Shah qaisar said:
7 years ago
P = i^2R.
I = 3mA= 0.003A,
R = 2000 * 1000/2000+1000= 666.6 ohm,
P = (0.003 * 0.003)* 666.6= 0.00599~6mW.
P =(.003 * .003).
Option C is correct.
I = 3mA= 0.003A,
R = 2000 * 1000/2000+1000= 666.6 ohm,
P = (0.003 * 0.003)* 666.6= 0.00599~6mW.
P =(.003 * .003).
Option C is correct.
(1)
Pijush patra said:
8 years ago
What is the total resistance of 5 kilo ohm, 5mege ohm connected in parallel?
Divya said:
9 years ago
P = VI -----> 1.
V = IR -----> 2.
Substitute above equation in 1,
We get P = I^2 * R.
R = 2*1/2 + 1,
R = 2/3,
P = (3*10^-3)^2*(2/3)*10^3 ohm,
p = 6mW.
V = IR -----> 2.
Substitute above equation in 1,
We get P = I^2 * R.
R = 2*1/2 + 1,
R = 2/3,
P = (3*10^-3)^2*(2/3)*10^3 ohm,
p = 6mW.
Krish said:
9 years ago
Two parallel resistor so equivalent resistor is 2/3 and also current 3 mA. So power loss is (p = Isquare R) 9 * 2/3 = 6 mA.
Chandrashekhar Manekar said:
9 years ago
Actually power( P) = I^2(R).
By using this, we can solve the problem.
By using this, we can solve the problem.
Addy said:
9 years ago
Simply,
Low resistance = Higher current flow.
So, we get applied voltage by dividing current in the ratio
2k/1k = 3mA.
1k = 2mA and 2k = 1mA.
Hence, applied voltage is 2V.
Now, we have I = 3mA, V = 2v.
P = V * I.
= 2 * 3mA.
= 6mA.
ANSWER 'C' IS CORRECT.
Low resistance = Higher current flow.
So, we get applied voltage by dividing current in the ratio
2k/1k = 3mA.
1k = 2mA and 2k = 1mA.
Hence, applied voltage is 2V.
Now, we have I = 3mA, V = 2v.
P = V * I.
= 2 * 3mA.
= 6mA.
ANSWER 'C' IS CORRECT.
Ana said:
10 years ago
1/R = 1/2+1/1.
1/R = 1+2/2.
R = 2/3 Kohm.
V = i*R.
V = 3 mA*2/3 Kohm.
V = 2 v.
P = V*I.
P = 2v*3 mA.
P = 6 mW.
1/R = 1+2/2.
R = 2/3 Kohm.
V = i*R.
V = 3 mA*2/3 Kohm.
V = 2 v.
P = V*I.
P = 2v*3 mA.
P = 6 mW.
Rashmi said:
1 decade ago
V = IR BY OHM'S LAW.
From the following data:
I = 3 m.
R1 = 1k.
R2 = 2k.
When resistors are connected in parallel.
Req is given by Req = R1*R2/R1+R2.
P = I^2*Req.
P = (3m)^2*(2/3)k.
P = 6 mW.
From the following data:
I = 3 m.
R1 = 1k.
R2 = 2k.
When resistors are connected in parallel.
Req is given by Req = R1*R2/R1+R2.
P = I^2*Req.
P = (3m)^2*(2/3)k.
P = 6 mW.
(1)
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