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< 2. For the circuit shown in Fig. 5.29 determine
(a) the reading on the ammeter, and (b) the
value of resistor R
[2.5 A, 2.5 S2]
692
А
ЗА
532
R
A
agement
Euced in
11.5 A4

Lt 2 For The Circuit Shown In Fig 529 Determine A The Reading On The Ammeter And B The Value Of Resistor R 25 A 25 S2 692 А ЗА 532 R A Agement Euced In 115 A4 class=

Sagot :

Jacob193idn

Answer:

Reading on the ammeter: [tex]2.5\; \rm A[/tex].

Value of [tex]R[/tex]: [tex]2.5\; \rm \Omega[/tex].

Explanation:

The three resistors are connected to the power supply in parallel. Hence, the voltage across each one of them would be equal to the voltage of the power supply.

Apply Ohm's Law to calculate the voltage across the resistor at the center:

Value of this resistor: [tex]5\; \rm \Omega[/tex].

Current through this resistor: [tex]I = 3\; \rm A[/tex]

By Ohm's Law, the voltage [tex]V[/tex] across this resistor would be [tex]5\; \rm \Omega \times 3\; \rm A = 15\; \rm V[/tex].

Hence, by the reasoning above, the voltage of the power supply and the voltage across the other two resistors would all be [tex]15\; \rm V[/tex].

Apply Ohm's Law (again) to calculate the current through the [tex]6\; \rm \Omega[/tex] resistor, given that the voltage across that resistor is [tex]15\; \rm V[/tex]:

[tex]\displaystyle I = \frac{V}{R} =\frac{15\; \rm V}{6\; \rm \Omega} = 2.5\; \rm A[/tex].

The ammeter is connected to the [tex]6\; \rm \Omega[/tex] resistor in a serial configuration. Hence, the reading of the ammeter would be equal to the current through this [tex]6\; \rm \Omega\![/tex] resistor: [tex]2.5\; \rm A[/tex].

Also because the three resistors are connected to the power supply in parallel, the current through the power supply would be equal to the sum of the current through each resistor.

Current through the power supply: [tex]11.5\; \rm A[/tex].

Current through the [tex]6\; \Omega[/tex] and the [tex]5\; \Omega[/tex] resistor: [tex]2.5\; \rm A[/tex] and [tex]3\; \rm A[/tex], respectively.

[tex]11.5\; \rm A = 2.5\; \rm A + 3\; \rm A + (\text{Current through $R$})[/tex].

Hence, the current through the unknown resistor [tex]R[/tex] would be:

[tex]11.5\; \rm A - 2.5\; \rm A - 3 \; \rm A = 6\; \rm A[/tex].

Apply Ohm's Law to find the value of resistor, given that the voltage across it is [tex]15\; \rm V[/tex] (same as the power supply) and that the current through it is [tex]6\; \rm A[/tex]:

[tex]\displaystyle R = \frac{V}{I} = \frac{15\; \rm V}{6\; \rm A} = 2.5\; \Omega[/tex].

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