Question

A student is provided with a 12.0-V battery of negligible internal resistance and four resistors with the following resistances: 100 Ohms, 30 Ohms, 20 Ohms, and 10 Ohms. The student also has plenty of wire of negligible resistance available to make connections as desired. 1. Using all of these components, draw a circuit diagram in which each resistor has nonzero current flowing through it, but in which the current from the battery is as small as possible. 2. Using all of these components, draw a circuit diagram in which each resistor has nonzero current flowing through it, but in which the current from the battery is as large as possible (without short circuiting the battery). 100 12V 20 w 30 100 The battery and resistors are now connected in the circuit shown above. Determine the following for this circuit.

Answer 1

1) using loo all these components, current through Battery will be minimum when resistance circuit is maximum I= V/R Maximum resistance when all resistors are in series in 2.0 12V 90 2 10 + Rmax = 100+ 30+20+ to 1602 3) i ( current through loe resistor) Imin = 12 160 0.075 need to find Reg of circuit Req too 30-2 o) 100 ohm 202 is in parallel with Serius Combinailcom 12v - of 20+ 30.12. tor Reg of this combination 1+1 heq to 2) when all resistors over all resistance and current will are in parallel will be minimum be manimum Req= 33.32 so Rea of chruit = 205 10S 100 3D Req= 5.20 12v 2.32 A Imax 12 5.2 This 33.3 e is in series with 1022 43,32 = 12 10.277 A 43.3 Power consumed in circuit Power = v*I V? R = 194 433 =/3.32 watt Energy Energy Power & time Power "I'm sec) 2 I²R (0.277) ²× 433 12x 0.277 And 3012 see = 50 minutes Croughly) Times 3072: Sec