two resistors of 25 ohms and 5 ohms respectively are connected in series to a 240V supply calculate the value of a third resistor to be connected in parallel with 25 ohms resistor so that the power dissipated shall be tripled
For each branch, P = E^2/R
Currently (no pun intended), we are dissipating 240^2/(25+5) = 1920 watts
So, we want to dissipate 3840 watts in the other branch.
240^2/R = 3840
R = 15 ohms
two resistors in series, 25,5 ohms.
power= 1920
want 240^2/(r'+5)=3840
figure r', then
1/r'=1/25+1/R
solve for R, the resistance going in parallel with the 25 ohm resistor
Go with bobpursley -- I misread the problem, connecting the parallel R across both series resistances.
P = E^2/(R1+R2) = 240^2/(25+5) = 1920 Watts = Current power.
Rt = (25+5)/3 = 10 Ohms = Total resistance required to triple the power.
Rt = R1*R3/(R1+R3) + R2 = 10.
(25*R3)/(25+R3) + 5 = 10
25R3/(25+R3) = 5
25R3 = 125 + 5R3
R3 = 6.25 Ohms.
To calculate the value of the third resistor, we need to determine the current passing through the series combination of the 25-ohm and 5-ohm resistors. Once we know the current, we can use it along with the desired power dissipation to find the resistance of the third resistor.
Let's start by calculating the total resistance of the series combination of the 25-ohm and 5-ohm resistors:
Total Resistance (R_total) = R1 + R2
= 25 ohms + 5 ohms
= 30 ohms
Next, we can find the current passing through the circuit using Ohm's Law:
Current (I) = Voltage (V) / Total Resistance (R_total)
= 240V / 30 ohms
= 8 A (Amperes)
Now, since we want the power dissipated to be tripled, we'll need to calculate the initial power dissipation and then determine the new desired power dissipation:
Initial Power (P_initial) = Current (I)^2 * Resistance (R1)
= 8 A^2 * 25 ohms
= 1600 W (Watts)
Desired Power (P_desired) = Initial Power (P_initial) * 3
= 1600 W * 3
= 4800 W
Now, to calculate the resistance of the third resistor (R3) needed to achieve the desired power dissipation, we can rearrange the formula for power dissipation:
Desired Power (P_desired) = Current (I)^2 * Resistance (R3)
Rearranging the formula gives:
Resistance (R3) = Desired Power (P_desired) / Current (I)^2
= 4800 W / (8 A)^2
= 75 ohms
Therefore, a third resistor of 75 ohms should be connected in parallel with the 25-ohm resistor to triple the power dissipated in the circuit.