Question

9.Two resistors of resistances 3 ohm and 6 ohm respectively are connected to a battery of 6V so as to have : a) Minimum resistance b)Minimum current
i) How will you connect the resistors in each case?
ii) Calculate the strength of the current in the circuit in both cases.
10.An electric bulb draws a current of 0.8A and works on 250V on an average 8 hours a day.If energy costs Rs 3 per kWh , calculate monthly bill for 30 days.

Answer 1

a) for minimum resistance, u connect both the resistors in parallel, the equivalent resistance would be:
\[((1/3)+(1/6))^{-1}\space = 1/2\space\Omega \]
the current I is therefore, V=IR, I=V/R, 6/0.5=12 A

b) For minimum current, we need to increase the resistance, we can do that if we put the resistors in series, then the equivalent resistance will be R=3+6=9, and current. I=V/R, I=6/9=1/3 A

10)
the power dissipated in the bulb can be found using the formula; P=IV, both quantities I and V are given, P=0.8(250) = 200 Watts

how much power is dissipated in this bulb if it runs for 8 hours a day for 30 days, well, just multiply: 200*8*30= 48000, which is equal to 48 kWh, now the rate for each kWh is given, 3 RS, 48*3= 144 RS/month

Answer 2

thank you so much for your help, onaogh.