The circuit diagram from the previous pane is reproduced at right for your conve

Question

The circuit diagram from the previous pane is reproduced at right for your convenience. e. Is bulb 2 brighter the, dimmer than, or as bright as bulb 1? Explain your reasoning. f. On the circuit diagram above, draw arrows to indicate the direction of the current through each battery and each bulb. Show that your answer is consistent with the student with whom you agreed in part c. g. Is the current through battery D greater than, less than, or equal to the current through battery C? Explain your reasoning. h. If bulb 1 were unscrewed, would the current through battery C increase, decrease, or remain the same? Consider the following incorrect student statement about circuit IX from the tutorial, shown at right: "When I apply Kirehhoff's loop rule when the switch is closed, the right loop has one battery and two bulbs a series. Thus the potential difference across the battery must equal the sum of potential difference across the bulbs, so each bulb will get half the potential difference of battery B and the - bulbs will be equally bright". Identify the flaw(s) in the student's reasoning.

Explanation / Answer

(e) bulb2 is brighter,

power = V^2 / R

voltage acrros bulb 1 = (Vx - Vz) ^ 2 / R

= V^2 / R

bulb 2 = (Vx - Vy)^2 / R

= (2v) ^ / R = 4 v^2 / R

so it clearly shows that power dissipiated across bulb2 is higher so brighter.

(f) use the KVL & KCL

V - i1×R = 0 ..........(1)

V - i2×R + V = 0

2V - i2×R = 0 ........(2)

i2 = 2V/R

I = i1 + i2 .........(3)

i1 = V/R

taking Vz = 0, Vx = V, Vy = -V

Vx > Vy > Vz

(g) current through battery C is greater than Current through battery D.

(h) if bulb1 removed, current through C would remain the same , while current through D would increase and it will be same as current through C.

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