Electronics - Parallel Circuits - Discussion
Discussion Forum : Parallel Circuits - General Questions (Q.No. 21)
21.
When a parallel 5 k
resistor and a 25 k
resistor have a 10 V supply, what is the total power loss?


Discussion:
13 comments Page 1 of 2.
Santhanakumari said:
2 years ago
Please clear the solution in detail. @Deepika.
p= 24 * 10^-3.
And P = 24 mwatts.
p= 24 * 10^-3.
And P = 24 mwatts.
Santhanakumari said:
2 years ago
Please clear the solution in detail. @Deepika.
p= 24 * 10^-3.
And P = 24 mwatts.
p= 24 * 10^-3.
And P = 24 mwatts.
Deepika.S said:
4 years ago
P=v*v/R.
R= (R1*R2) / (R1+R2).
R= (5*25) / (5+25).
R=4.16*10^3ohms.
P= (10*10) / (4.16*10^3).
P=24*10^-3.
P=24mwatts.
R= (R1*R2) / (R1+R2).
R= (5*25) / (5+25).
R=4.16*10^3ohms.
P= (10*10) / (4.16*10^3).
P=24*10^-3.
P=24mwatts.
Mani said:
5 years ago
We know p=v*v/R.
Given R1&R2 are connected in parallel,
R1 = 25k&R2=5k,
R = R1llR2,
R = R1*R2/R1+R2=4.16kohms,
P = 10*10/4.16*10^3,
P = 24mwatts.
Given R1&R2 are connected in parallel,
R1 = 25k&R2=5k,
R = R1llR2,
R = R1*R2/R1+R2=4.16kohms,
P = 10*10/4.16*10^3,
P = 24mwatts.
Prashant said:
6 years ago
R1 =5K THATS WHY P1 =10*2/5K =20mw (P=V2/R).
R2= 25K THATS WHY P2 = 10*2/25K =4mw.(P=V2/R).
SO,
P = P1+P2 =24mw.
R2= 25K THATS WHY P2 = 10*2/25K =4mw.(P=V2/R).
SO,
P = P1+P2 =24mw.
ANITA NAGAR said:
1 decade ago
RT = R1*R2\(R1+R2) = 4.166.
P = V^2\R = 24.03MW.
P = V^2\R = 24.03MW.
Shahid said:
1 decade ago
@Jayashree.
Both the resistors are parallel to each other means (25*5) / (25+5) =4. 16.
Both the resistors are parallel to each other means (25*5) / (25+5) =4. 16.
Viju said:
1 decade ago
Rt=25*5/30.
=5/6 kohm.
P=V*V/R.
=100*6/5.
=24mW.
=5/6 kohm.
P=V*V/R.
=100*6/5.
=24mW.
Abi said:
1 decade ago
P=VI
BUT I=V/R
SO P=V.V/R
R=R1.R2/R1+R2
BCOZ RESISTORS ARE PARALLEL HERE.
R=4.166K ohms.
P=100/4.166K
=24mw.
BUT I=V/R
SO P=V.V/R
R=R1.R2/R1+R2
BCOZ RESISTORS ARE PARALLEL HERE.
R=4.166K ohms.
P=100/4.166K
=24mw.
Chinnu said:
1 decade ago
Since, r1||r2
rt = 4.166
P = V*V/R
=> P=24.03mW
rt = 4.166
P = V*V/R
=> P=24.03mW
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