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