Electronic Principles

Deleted member 91840 · Nov 5, 2015

Hi,
I have a module on electronic principles. One of the questions is
Three resistors of 50 ohms,50 ohms & 5 ohms are connected in series across a voltage source (Battery). If the power dissipated in the 50 ohms resistor is 100w, calculate the voltage drop across each resistor and the total voltage that must be applied to the series circuit.
I wasn't sure if i was to assume because its a battery its 12v or to work out from my two known values the resistor and the Watts so I worked out my voltage from the resistor value and the watts. My working out is as followed. I am not sure if I am on the right track or the complete wrong one any help would be appreciated?

I=70.71/105= 0.67MA

V=I*R
V1= 0.67* 50= 33.5V
V1= 0.67* 50= 33.5V
V3= 0.67*5=3.35V
V1+V2+V3= 70.35 V

Richard Burns · Nov 5, 2015

Current in a series circuit is the same at all points in the circuit.
You have P = V²/R so √(PR) = V
This is the voltage dropped on the 50 Ω resistor (70.71 V) so the current is relating to the 50Ω resistor.
i.e. what current will flow to drop 70.71 V over a 50 Ω resistor
i.e. I= V/R = 70.71/50 = 1.4342 A
Then do the same calculation you have done using the new current and add up as you have done to get the total voltage dropped over allt he resistors and so this is the supply voltage needed to get 100W dissipated in the 50Ω resistor.

However I am very rusty on this type of thing and so may be completely wrong. It just seems right.

telectrix · Nov 6, 2015

rusty or not. looks right to me, richard.

Lewisham · Jan 31, 2016

j10nba said:
Hi,
I have a module on electronic principles. One of the questions is
Three resistors of 50 ohms,50 ohms & 5 ohms are connected in series across a voltage source (Battery). If the power dissipated in the 50 ohms resistor is 100w, calculate the voltage drop across each resistor and the total voltage that must be applied to the series circuit.
I wasn't sure if i was to assume because its a battery its 12v or to work out from my two known values the resistor and the Watts so I worked out my voltage from the resistor value and the watts. My working out is as followed. I am not sure if I am on the right track or the complete wrong one any help would be appreciated?

I=70.71/105= 0.67MA

V=I*R
V1= 0.67* 50= 33.5V
V1= 0.67* 50= 33.5V
V3= 0.67*5=3.35V
V1+V2+V3= 70.35 V

I=70.71/105= 0.6; This is wrong. The Current at the Series resistors are same, but the voltage drops(potential difference) are different. So, 70.71 is voltage drop only at the resistor R1.

To find current(I):
I=V1/R1, 70.71/50
I=1.414A - This current is the same to all resistors.

Since we have current is easy to calculate V2 and V3, but the voltage drop at R2 is the same as R1 because they have the same resistivity capacity.
V3=1.414 * 5
V3=7.071

There are two ways to calculate total voltage by adding V1,V2,V3 or by V=IR, 1.414 * 105 = 148.47.

THE ANSWERS:
V1=70.71, V2=70.71 and V3=7.071.
Total voltage needed = 148.47