Isaac Physics: A Power Problem

Assuming you are familiar with Ohm's law ($V = IR$) and the equation for power dissipated in resistor ($P = I^{2}R$), then you can split the problem up as follows.

First, work out the resistances of each of the three sub-blocks of the circuit (first block being the $2~\Omega$ resistors in parallel, second block being the $2~\Omega$ resistor on its own, and third block being the three sets of resistors in parallel. Then, work out the current through each block in terms of the (unknown) total voltage $V$ and the resistances you just calcuated (hint: the total current should be equal to $V/9$).

Then, you can work out the power dissipated in each resistor for every block, set it equal to $2~W$ and see which gives the lowest limiting voltage. The first two blocks are straightforward, but the last block is a bit trickier. Remember that the current through each branch of block must sum to the total current through the block, and don't forget to split the current going through the two $20~\Omega$ resistors in parallel.

See how you go with that.

Ok, so do I just do the calculation for each block as (V/Rtotal for that block)2 multiplied by Rtotal for that block = 2W?

Yes, although the power equation is actually $V^{2}/R$. I actually think it may be easier to apply it in the form $I^2 R$ but use whatever form works best for you.

Let's look at the first block (two $2~\Omega$ resistors in parallel) as an example. The resistance of this block is $(1/2 + 1/2)^{-1} = 1~\Omega$, using formula for resistors in parallel, and the current flowing through each resistor is half of the total current $1/2 \times V/9 = V/18$ (you should be able to calculate total current by applying Ohm's law to whole circuit). So, the power dissipated in each of the two resistors is $(V/18)^2 \times 2 = V^2/162$. This must be less than 2 W, and so $V^2 < 234 \implies V < 18$ volts.

See if you can apply the same procedure to the other blocks.

Ok, so I did that for the second block which is just the single 2ohm resistor and found that V<9volts

Correct.



You're right - it isn't. The $15~\Omega$ resistor will carry a greater share of the current than the other two branches. To work out the current through each branch, consider the potential difference across each branch ($6 V/9$) and then use the power formula as before.

Assuming you are familiar with Ohm's law ($V = IR$) and the equation for power dissipated in resistor ($P = I^{2}R$), then you can split the problem up as follows.

First, work out the resistances of each of the three sub-blocks of the circuit (first block being the $2~\Omega$ resistors in parallel, second block being the $2~\Omega$ resistor on its own, and third block being the three sets of resistors in parallel. Then, work out the current through each block in terms of the (unknown) total voltage $V$ and the resistances you just calcuated (hint: the total current should be equal to $V/9$).

Then, you can work out the power dissipated in each resistor for every block, set it equal to $2~W$ and see which gives the lowest limiting voltage. The first two blocks are straightforward, but the last block is a bit trickier. Remember that the current through each branch of block must sum to the total current through the block, and don't forget to split the current going through the two $20~\Omega$ resistors in parallel.

See how you go with that.