Physics Task: Circuits
1. Determine the reading of the voltmeter shown in the figure. Consider the values of the
resistors as: R1 = 5Ω, R2 = 4.4Ω, R3 = 7.7Ω, R4 = 4.5Ω, R5 = 4.8Ω and that the source
provides a voltage V = 8.2V .
V
8.2V
8.2 V
=
=
=0.31 A
R Total ❑ 5 Ω+ 4.4 Ω+7.7 Ω+ 4.5 Ω+4.8 Ω 26.4 Ω
2. Let us be two cylindrical conductors connected in parallel, to which a potential difference
of V = 170V is applied. The two conductors are made of the same material, but the first is
I=
6 times the length of the second, and the radius of the second. The resistance of the
second is R2 = 469Ω. Determine the equivalent resistance.
Let ρ is the resistivity of material second resistance R2=469Ω
ρl
=469
π R2 ❑❑❑
ρ×(l ×6)
π ¿¿
1
(
) × R1=469
150
R1=469 ×150=70350 Ω
The equivalent resistance of parallel resistors
R × R 2 70350 × 469
Req = 1
=
=465.894 Ω
R1 + R2 70350 × 469
3. In the circuit shown in the figure, the ideal ammeter measures I = 0.4A in the indicated
sense, and the ideal voltmeter measures a potential drop of V = 8.8V passing from b a a.
Determines the value of the emf ε2.
Data: R1 = 56.2Ω, R2 = 23.3Ω, R3 = 27.4Ω.
1
R1=56.2 o h m
R2=23.3 o h m
R3=27.4 o h m
ε 2=R2 (I −I 1)−R3 I =0
V =I 1 R 1
I 1=
8.8
=0.156 A
56.2
ε 2−23.3(0.4−0.156)−27.4(0.4 )=0
ε 2=5.685+10.96=16.645 V