Problem 4 The diagram to the right shows a 6.0-V battery connected to an ammeter

Question

Problem 4 The diagram to the right shows a 6.0-V battery connected to an ammeter and a network of four resistors, each with a resis- tance of 240 2. Don't worry about the fact that the resistors are shown as rectangles instead of the normal zig zags: this is how they are shown in European diagrams. I promise that they are still really good resistors. (a) What is the equivalent resistance for this network of four resistors? pts (b) What current does the ammeter read? pts (c) (Low points/work ratio!) What is the potential drop across the upper right resistor? [i ptl Problem continues on neat page.

Explanation / Answer

(a)

Upper two resistor are in series.

Rs = 240 + 240 = 480 ohm

this is in parallel with 240 ohm resister.

Rp = 480*240 / (240 + 480) = 160 ohm

This is in series with 240 ohm. so equivalent resistace of circuit-

Req = 160 + 240 = 400 ohm

(b)

current in ammeter,

l = V / Req = 6 / 600

l = 0.01 A

(c)

V = l*R

so, first we will calculate current through upper right resistor.

At parallel combination current will divide in opposite ratio of resistance.

current in upper branch,

l' = 240*0.01 / (240 + 480) = 0.003 A

so, potential drop in upper right resistor is,

V' = l'*R

V' = 0.003 * 240

V' = 0.8 V

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