Test: Current and Resistance - MCAT MCQ

Electrical current is defined as charge flow, or in mathematical terms, charge transferred per time: I = Q/Δt. 
A 15 A current that acts for 0.1 s will transfer 15 A × 0.1 s = 1.5 C of charge.

The resistance of a resistor is given by the formula R = ρL/A 
Thus, there is a direct proportionality between resistance and resistivity. Because the other variables are equal between the two resistors, we can determine that if R₁:R₂ is a 1:2 ratio, then ρ₁: ρ₂ is also a 1:2 ratio.

We are told that transformers conserve energy so that the output power equals the input power. Thus P_out = P_in, or I_outV_out = I_inV_in. There is therefore an inverse proportionality between current and voltage. If the output voltage is 300% of the input voltage (3 times its amount), then the output current must be 1/3 of the input voltage. This can be represented as a 1:3 ratio.

To determine the moles of charge that pass through the circuit over a period of 10 s, we will have to calculate the amount of charge running through the circuit. Charge is simply current times time, and the current can be calculated using Ohm's law:

Then, calculate the number of moles of charge that this represents by using the Faraday constant and approximating F as

This is closest to choice (A).

This question should bring to mind the equation 
where ε₀ is the permittivity of free space, A is the area of the plates, and d is the distance between the plates. From this equation, we can infer that doubling the area will double the capacitance, and halving the distance will also double the capacitance. Therefore, the new capacitance is four times larger than the initial capacitance.

Power is energy dissipated per unit time; therefore, the energy dissipated is E = PΔt. In the five-second interval during which the resistor is active, it has a 2 A current for three of those seconds. The power dissipated by a resistor R carrying a current I is P = I²R. Therefore, the energy dissipated is

E = I²RΔt =(2 A)²(10 ?)(3 s)= 4 × 10 × 3 = 120 J

The electric field between two plates of a parallel plate capacitor is related to the potential difference between the plates of the capacitor and the distance between the plates, as shown in the formula E = V/d.
The addition of another battery will increase the total voltage applied to the circuit, which, consequently, is likely to increase the electric field. The addition of a resistor in series will increase the resistance and decrease the voltage applied to the capacitor, eliminating choice (A). Adding a resistor in parallel will not change the voltage drop across the capacitor and should not change the electric field, eliminating choice (B). Increasing the distance between the plates, choice (C), would decrease the electric field, not increase it.

While this is primarily a recall question, it should also be intuitive. Voltmeters are attempting to determine a change in potential from one point to another. To do this, they should not provide an alternate route for charge flow and should therefore have infinite resistance. Ammeters attempt to determine the flow of charge at a single point and should not contribute to the resistance of a series circuit; therefore, they should have no resistance.

Kirchhoff's loop rule states that the total potential difference around any closed loop of a circuit is 0 V. Another way of saying this is that the voltage gained in the battery (6 V) will be used up through the resistors. Because this charge both started and ended at the positive terminal, its total potential difference is therefore 0 V. 6 V, choice (D), is the voltage gained in the battery as well as the voltage drop in the resistors—creating a net sum of 0 V.

This question tests our understanding of batteries in a circuit. The voltage across the terminals of the battery when there is no current flowing is referred to as the electromotive force (emf or ε of the battery). However, when a current is flowing through the circuit, the voltage across the terminals of the battery is decreased by an amount equal to the current multiplied by the internal resistance of the battery. Mathematically, this is given by the equation
V = ε – ir_int
To determine the emf of the battery, first calculate the voltage across the battery when the current is flowing. For this, we can use Ohm's law:
V = IR
= (0.5 A) (3Ω) = 1.5 V
Because we know the internal resistance of the battery, the current, and the voltage, we can calculate the emf:
ε = V + ir_int  
= 1.5 V + (0.5 A) (0.1Ω)
= 1.5 + 0.05 = 1.55 V
The answer makes sense in the context of a real battery because its internal resistance is supposed to be very small so that the voltage provided to the circuit is as close as possible to the emf of the cell when there is no current running.