Question

1. Consider the circuit below consisting of a battery, identical light bulbs, and an uncharged capacitor. a) Discuss the brightness of the bulbs as you close the switch and watch for a few seconds. A B HE D с

Answer 1

Let's say resistence 'R'as each . bulb has they are are Identical, circuit So V mo can be drauon Lue (initially capaciton act as a get the To short circuit) brightness of the bulbs, we need to find the voltage across them. formula Power (P) = higher the voltage a bulb gives higher brightness. say current through circuit = I amp from v/ shows that across let then, A NH R voltage across VA IR R Ч х R 70 voltage across X4 -- in Vxy = [( Rc+RD) 11R6] I Uxy = I (PR11R) he {Req= 28 8 and (C+D) are parallel, so ar R \\xy = 2 IR Rea } As vxy = UB ŽIR and C and D are Holtage across ĶIR vxy Lice. Vc = 2 VD= ŠIR from above calculation, we NAS VB > Ve=UD found that as pau?

Means Bult A has highest brightness has brighness higher than c and D but Bulb & lower than bulb A, Bulb c and Bulb D lower than A and B. have equal brightness but 2 After several seconds, capacitor also provide some resistance and these fose has some voltage drop across it again calculating the voltage drop across say (vo) bulbs. VA E IR is short circuit due to this additional Bulb c wire So Ve=0 Novo circuit will look like an Vxy = I ((R+R)118 ] RO B Ht Uxy < IR/2 X y D xxy = IR (If there is No capacitor) and UD = vxy So VDL IR IR VB \xy vo so VB < IR HR-V vo NA > Vo> VB lie and Ve=0 Means Bulb a has highest brightness. lauseest brightness Bulb D has has brightness higher than B out lower Bulb B has than A Bulb c has no light and No brightness,