Answer:
[tex]20W[/tex]
Explanation:
The first thing to bear in mind is that the 10 ohms and 20 ohms resistors are in parallel; and when resistors are in parallel the voltage drop across them will be the same.
Therefore if we can obtain the voltage drop across the 10 ohms resistor, then we can apply the same voltage to the 20 ohms resistor (since it is in parallel with the 10 ohms resistor) and then use that same voltage to obtain the power dissipated therein.
The following are the relationships between power (p), current (I), resistance (R) and voltage (V);
[tex]V=IR.............(1)\\P=IV..............(2)\\[/tex]
Substituting equation(1) into (2), we obtain the following;
[tex]P=I(IR)\\P=I^2R............(3)[/tex]
Let the power across the 10 ohms resistor be [tex]P_{1[/tex] and that of the 20 ohms resistor be [tex]P_{2[/tex]. Also their their respective currents and voltages be [tex]I_1[/tex], [tex]I_2[/tex], [tex]V_1[/tex] and [tex]V_2[/tex].
Given;
[tex]R_1=10ohms; P_1=40W[/tex]; [tex]R_2=20ohms; P_2=?[/tex]
Using equation (3), we can obtain [tex]I_1[/tex] as follows;
[tex]P_1=I_1^2R_1\\40=I_1^2*10\\therfore\\I_1^2=40/10=4\\I_1=\sqrt{4}=2A[/tex]
We now use this to obtain the voltage drop across [tex]R_1[/tex] as follows;
[tex]V_1=I_1R_1\\V_1=2*10=20V[/tex]
Recall we said that the voltage drop across the 10 ohms and the 20 ohms resistors are the same because they are in parallel, therefore [tex]V_1=V_2=20V[/tex].
We now use this value to determine the current in [tex]R_2[/tex] As follows through the relationship defined in equation(1);
[tex]V_2=I_2R_2\\therefore\\I_2=V_2/R_2\\I_2=20/20=1A[/tex]
Finally we now use equation (2) to determine [tex]P_2[/tex] as follows
[tex]P_2=I_2V_2\\P_2=1*20\\P_2=20W[/tex]
Note please: This is not the only method that could be used in solving this problem. There are other approaches that could be used, but I feel the approach used here should be simplified and straight forward to understand. Thanks.
