Question

7. Three resistors having resistances of 2.00 2, 3.002, and 4.002 are connected in series to a 24.0-Vbattery that has negligi- ble internal resistance. Find a) the equivalent resistance of the combination; b) the current in each resistor; c) the total cur rent through the battery; d) the voltage across each resistor, e) the power dissipated in each resistor.

Answer 1

(a)since the three resistors are in series,total resistance=2+3+4=9 ohms

(c) current in the combination =total emf/total resistance=24/9= 2.66 amperes

(b) current in each resistor will be same as they are in series and this current will be 2.66 amperes

(d) voltage across 2 ohms resistor= current * resistance=2* 2.66= 5.32 volts

voltage across 3 ohms resistor=3*2.66=7.98 volts

voltage across 4 ohms resistor=4*2.66=10.64 volts

(e) power across 2 ohms resistor=VI=2.66*5.32=14.1512 watts

power across 3 ohms resistor=2.66*7.98=21.2268 watts

power across 4 ohsm resistor-2.66*10.64=28.3024 watts