This physics problem tests your understanding of fundamental physical laws and their applications. The step-by-step solution below breaks down the problem using relevant equations and physical reasoning.
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here are the solutions to the questions you've provided: 4. The resistance of a 120 V light bulb that draws a current of 1.25 A is ________. Step 1: According to Ohm's Law, the resistance R is given by: R = (V)/(I) where V is the voltage and I is the current. Step 2: Substitute the given values: V = 120 \, V and I = 1.25 \, A. R = 120 \, V1.25 \, A R = 96 \, The correct option is B. 96 \, 5. A certain wire has a resistance R. Another wire, exactly the same except having twice the diameter, has the resistance ________. Step 1: The resistance R of a wire is given by the formula: R = (L)/(A) where is the resistivity of the material, L is the length, and A is the cross-sectional area. Step 2: The cross-sectional area A of a circular wire is given by A = r^2, where r is the radius. Since diameter d = 2r, then r = d/2, so A = ((d)/(2))^2 = ( d^2)/(4). Step 3: Substitute the area into the resistance formula: R = (L)/( d^2)4 = (4 L)/( d^2) Step 4: For the second wire, the diameter is d' = 2d. All other parameters (, L) are the same. Let its resistance be R'. R' = (4 L)/( (d')^2) = (4 L)/( (2d)^2) = (4 L)/( (4d^2)) = (1)/(4) ( (4 L)/( d^2) ) Step 5: From Step 3, we know that R = (4 L)/( d^2). So, R' = (1)/(4) R The correct option is D. (1)/(4)R 6. The current in the 5 Ω resistor is ________. Step 1: The resistors 5 \, and 10 \, are connected in series. The total equivalent resistance R_eq for series resistors is the sum of individual resistances: R_eq = R_1 + R_2 R_eq = 5 \, + 10 \, = 15 \, Step 2: A potential difference V = 60 \, V is applied across the series combination. According to Ohm's Law, the total current I flowing through the circuit is: I = (V)/(R_eq) I = 60 \, V15 \, I = 4 \, A Step 3: In a series circuit, the current is the same through all components. Therefore, the current in the 5 \, resistor is 4 \, A. The correct option is A. 4 A 7. The potential difference across the 5 Ω resistor is ________. Step 1: From the previous question (Question 6), we found that the current I flowing through the series circuit is 4 \, A. Step 2: To find the potential difference V_1 across the 5 \, resistor (R_1), we use Ohm's Law: V_1 = I × R_1 Step 3: Substitute the values: I = 4 \, A and R_1 = 5 \, . V_1 = 4 \, A × 5 \, V_1 = 20 \, V The correct option is A. 20 V 8. A current of 9 A flows when a 120 V battery is connected across a 12 Ω resistor. The battery has an internal resistance of ________. Step 1: For a circuit with a battery having an electromotive force (EMF) E and internal resistance r, connected to an external resistor R, the current I flowing in the circuit is given by: I = ER + r Step 2: We are given: I = 9 \, A, E = 120 \, V, and R = 12 \, . We need to find r. Rearrange the formula to solve for r: I(R + r) = E R + r = EI r = EI - R Step 3: Substitute the values: r = 120 \, V9 \, A - 12 \, r = 13.333... \, - 12 \, r = 1.333... \, Rounding to one decimal place, r ≈ 1.3 \, . The correct option is A. 1.3 \, 9. A 10 Ω resistor and a 5 Ω resistor are connected in parallel. Their equivalent resistance is ________. Step 1: For two resistors connected in parallel, the equivalent resistance R_eq is given by the formula: (1)/(R_eq) = (1)/(R_1) + (1)/(R_2) Step 2: Substitute the given values: R_1 = 10 \, and R_2 = 5 \, . (1)/(R_eq) = (1)/(10 \, ) + (1)/(5 \, ) Step 3: Find a common denominator and add the fractions: (1)/(R_eq) = (1)/(10 \, ) + (2)/(10 \, ) (1)/(R_eq) = (3)/(10 \, ) Step 4: Invert the result to find R_eq: R_eq = (10)/(3) \, R_eq ≈ 3.33 \, The correct option is B. 3.3 \, Got more? Send 'em!