Question

I just did a lab and I can't figure out why the voltage behaves this way.

Circuit one: A 1.5 volt battery and a voltmeter V= 1.319V

Circuit two: Same as circuit one but there is a bulb parallel with the voltmeter and the voltage drops to 1.156V. Shouldn't the voltage be the same because they are in parallel and how can you use theory or otherwise to explain this? Is it due to resistance decreasing because the bulb abd voltmeter are in parallel?

Circuit 3. A second bulb is added in parallel so now there is three things in parallel, bulb, bulb, and voltmeter and the voltage drops again to 0.990V why? Does the current in all three circuits stay the same? Isn't the voltage supposed to be the same because it is the voltage of the battery?

If you are given a circuit in with a 1.5 V battery connected to an ammeter is series and two things in parallel--> a light bulb and a voltmeter and the voltage is 1.69V and the current is 197.46 micro amps could this be the current flowing through circuits one through three?

Is there a way to explain this voltage drop quantitativitly?

Answer 1

in circuit 2 as due to bulb total resistance would have droped and hence voltage has dropped as the combination of restance ofvoltmeter as well as the the resistance of bulb will decrease and it should be in series with some resistor to complete the circuit. As we know that in series connection volatge access lower resistance is less. Now due to addition of bulb resistance of arm containing voltmeter as well as bulb is decreased and the resiatance of rest of the circuit is still same hence voltage across voltmeter will be less as compare to circuit 1

the above explanation also stand true for circuit 3.

no this current can not be same for all circuit as in all three circuit value of resistance changes and hence value of current will also change.

voltage drop = i×R

here you can calculate voltage drop across any resistor if you know the value of current