Resistance in Circuits

A student wants to design a circuit containing a 6V battery and wants to use 100 ohm resistor to limit the current in the circuit. He only has 100 ohm resistors which have a power rating of 0.25W.

(a) Show that the p.d. of 6V connected across a 100 ohm resistor will generate more power than 0.25W

I have done this using P = V^2/R and got 0.36 W which is > 0.25W

(b) What is the least number of identical resistors which the student can use in order to provide the same total resistance without exceeding the power limit for each?

Not sure how to go about this or what it is quite asking for. What does it mean by the "same total resistance"? Same as what?

(c) Draw a circuit to show how they would be connected to the battery?

Not sure either! Guessing you will need to know answer to part b first though


Any help would be much appreciated!

How can you arrange a number of 100 ohm resistors, in parallel, series, or a combination such that the effective resistance across the network is 100ohms and the power dissipated in EACH resistor is less than or equal to 0.25W.

Oh I see. Would you have 2 x 100 ohm resistors in parallel and then another 2 x 100 ohm resistors also in parallel but in series with the first set (if that makes sense). This would give the effective resistance of each pair as 50 ohms, each giving the required 100 ohm total. The power consumption will then be less than 0.25W - it will be 0.09 W as it will go down by a factor of 4 since the power has halved and P = V^2/R. Is that correct reasoning? I can't see how you can get the desired outcome with less than 4 resistors so I think that's right.

Exactly, correct.

Thanks for the help.

Oh I see. Would you have 2 x 100 ohm resistors in parallel and then another 2 x 100 ohm resistors also in parallel but in series with the first set (if that makes sense). This would give the effective resistance of each pair as 50 ohms, each giving the required 100 ohm total. The power consumption will then be less than 0.25W - it will be 0.09 W as it will go down by a factor of 4 since the power has halved and P = V^2/R. Is that correct reasoning? I can't see how you can get the desired outcome with less than 4 resistors so I think that's right.