Circuits with Series and Parallel Wiring

Snape1830

Oh, right yeah, I just didn't put the decimal point in. Silly me.

Thanks!

Rajatmo

questions like these can be solved in your head!
first, for the current, the net resistance of the topmost two resistors is 5 ohm. now, the reciprocal of 5 is .2 and the reciprocal of 20 is .05 as you should be able to do immediately. they add up to be .25 - which is the reciprocal of the net resistance of the equivalent resistor containing the 20 ohm resistor and also the reciprocal of 4. so, to date, the net resistance is 4 ohms and this adds to the 6 ohms resistor in series. now, clearly, the net current through the battery is 20/(6+4) A = 2 A. this 2 A current divides in two sections in the ratio 4:1 (the current varies inversely as the resistance with p.d. constant). so, the current going through the ammeter is 2[itex]\times[/itex]4/5 = 16/10 = 1.6 A.
for the power, the current 1.6 A just divides in two equal parts, so it's .8 A through each 10 A resistor. so, the power dissipated is 10[itex]\times[/itex].8² = 6.4 W.

so, don't use the loop and the junction rules unless you really need them. they'll mess up everything. try to solve this kinds of problems intuitively or using the method i've described - they are really easy.