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 .
Let R(23) be the parallel combination of R2 and R3. Therefore
1/ R(23) = 1/ R2 + 1/R3
= 1/4.4 Ω + 1/7.7 Ω
= 12.1/33.88 Ω (get the reciprocal)
R(23) = 33.88 Ω/12.1
= 2.8 Ω
Let R(234) be the series combination of R(23) and R4. Therefore
R(234) = R(23) + R4
= 2.8 Ω + 4.5 Ω
= 7.3 Ω
Let R(2345) be the parallel combination of R(234) and R5. Therefore
1/R(2345) = 1/R(234) + 1/R5
= 1/7.3 Ω + 1/4.8 Ω
= 12.1/35.04 Ω (get the reciprocal)
R(2345) = 35.04 Ω/12.1
= 2.9 Ω
Let RT be the series combination of R1 and R(2345). Therefore
RT = R1 + R(2345)
= 5 Ω + 2.9 Ω
= 7.9 Ω
Let IT be the current flowing along RT. Using Ohm’s law, we can solve it by
IT = VT/RT
= 8.2 V/7.9 Ω
= 1.04 A
But IT = I1 = I(2345) = 1.0 A (since current is the same along the resistors in series)
V1 = I1R1
= (1.04 A) (5 Ω)
= 5.2 V
V(2345) = I(2345)R(2345)
= (1.04 A) (2.9 Ω)
= 3.0 V
But V(2345) = V(234) = V5 = 3.0 V (since voltage is the same across the resistors in parallel).
Therefore the reading in voltmeter across R5 is 3.0 V.
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
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.
Resistance can be calculated by R = ρL/A where ρ, L and A is the resistivity, length of
the conductor and cross-sectional area, respectively.
Let R1, ρ1, L1 and A1 be the resistance, resistivity, length, and area of the first conductor,
respectively while R2, ρ2, L2 and A2 be for the second conductor. Therefore, we can write an
equation for the resistances, R1 and R2: R1 = ρ1L1/A1 and R2 = ρ2L2/A2. However the two
conductors are made of the same material which means they have the same resistivity: ρ1 =
ρ2. Take note also of the conditions “the first is 6 times the length of the second, and 1/5 the
radius of the second”. Therefore L1 = 6L2 ⇔ L2 = L1/6 and r1 = r2/5 ⇔ r2 = 5r1.
From R1 = ρ1L1/A1 and R2 = ρ2L2/A2, we can rewrite the equations for ρ1 and for ρ2,
ρ1= R1A1/L1
ρ2= R2A2/L2
But ρ1= ρ2,
R1A1/L1 = R2A2/L2
Following the conditions L1 = 6L2 and r1 = r2/5 ⇔ r2 = 5r1, and substitute these into the
equation and applying the formula for the cross-sectional area of a cylinder (A = πr2)
R1(πr12)/L1 = R2(πr22)/L2
R1((π(r2/5)2)/6L2 = R2(πr22)/L2
R1((π(r22/25))/6L2 = R2(πr22)/L2 (Quantities π, r22 and L2 will be cancelled out)
R1/150 = R2 (multiply both side by 150 and substitute the value of R2)
R1 = 150 (469 Ω)
R1 = 70,350 Ω
Applying the formula of equivalent resistance (REQV) for parallel connection,
1/REQV = 1/R1 + 1/R2
= 1/70,350 Ω + 1/469 Ω
= 70,189/32,994,150 Ω
REQV = 32,994,150 Ω/70,189
= 470.08 Ω
Therefore, the total resistance (equivalent resistance) of the two conductors is about 470 Ω.
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.
We can solve this by Kirchoff’s laws. We have two loops here. Let the blue loop be L1 and the red is L2. We also
designated one node – node x.
x
L1: - ε1 + ε2 – 8.8 V = 0
L2: ε2 – (0.4A) (27.4 Ω) – I2(23.3 Ω) = 0
At node X,
0.4 A + (8.8 V/56.2 Ω) = I2(23.3 Ω)
0.4 A + 0.16 A = I2(23.3 Ω)
0.56 A = I2(23.3 Ω) (divide both side by 23.3)
0.024 A = I2
Substitute this value to the equation in L2, we have,
ε2 – (0.4A) (27.4 Ω) – (0.024 A) (23.3 Ω) = 0
ε2 – 10.96 V – 0.56 V = 0
ε2 = 11.52 V