8-5: Ohm\'s LawPost-Lab Questions Date: Marks: Max = 3+3+2+2 = 10 Name: 1. In th

Question

8-5: Ohm's LawPost-Lab Questions Date: Marks: Max = 3+3+2+2 = 10 Name: 1. In the circuit shown, what are the values of the current Ii. lh and lb 10 5 0 10 13 12 V in the crcuit shown in Q1, if the negative probe of the voltmeter is placed at point marked A',what will be the reading when the other probe is placed at points: 2· 3. You have one battery and two lamps. Will the total light output from both lamps be more when they are connected to the battery in series or parallel, and why? 4. You have two batteries and one lamp. Will the lamp be brighter when the batteries are connected in series or parallel, and why?

Explanation / Answer

1) by ohms law, i1 = V/Req = 12/(5+10/2)

= 1.2 A

i2 and i3 be equal as both have equal resistance.

i2 = 0.5*i1 = 0.6 A

i3 = 0.6A

2) B= 12V

D = 12V

E = 12 - i2*R2 = 12-10*0.6 = 6 V

3)Light output will be more when they will be in parallel, because potential across each bulbs will be more in case of parallel, so more current will pass.

4) It will be brighter when they are connected in series. Because emf i.e potential difference across bulb will be twice in case of series.

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