Electronics - Parallel Circuits - Discussion

Discussion Forum : Parallel Circuits - General Questions (Q.No. 4)

Parallel Circuits

What is the total power loss if 2 k

and 1 k

parallel-connected resistors have an I_T of 3 mA?

6 mW

36 mW

Answer: Option

Explanation:

No answer description is available. Let's discuss.

Discussion:

39 comments Page 3 of 4.

Prasanna said: 1 decade ago

Ramprabhu is wrong,
Resistance value here given is Kohms and he consider it to be only ohms

Irshad k a said: 1 decade ago

R = 2/2+1/2.
R = 3/2.
R = 2/3.
R = 0.6668.

P = I2R.
P = 3*3*0.6668.
P = 6(approximately).

Naseem Ul Aziz said: 1 decade ago

p=vi
p1=?
p2=?
pt=?
p1=????
vt=ir
rt=666.666 ohm
vt=2 volt
p1=2m w
p2=4mw
pt=p1+p2=6 m w

U.suruthi said: 1 decade ago

R = (2*1)/(2+1) = 2/3.

I = 3*10^-3.

P = I^2*R.

P = (3*10^-3)^2*(2/3).

P = 6*10^-3w.

Sree said: 1 decade ago

The formula for resistor when 2 resistors are connected in parallel is 1/R=1/R1+1/R2.

Linton said: 1 decade ago

@piyush suthar

How 1/2 + 1/1 will makes 2/3 ?

1/2 = (.5)+(1) = 1.5 = 3/2 not 2/3.

Mamatha.m said: 1 decade ago

WKT,
P=I2*R
where Rp=R1R2/R1+R2
Rp=(2*1)/(2+1)
Rp=2/3k ohm
P=(3m)^2*(2/3)k
P=6mW

Abc said: 1 decade ago

What would be the total power loss had the circuit been connected in series?

Pijush patra said: 8 years ago

What is the total resistance of 5 kilo ohm, 5mege ohm connected in parallel?

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)

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