First we calculate the current across each resistor. From Ohm's law, we arrive at a general formula for current: Plugging in values, the current across the resistor R1 is equal to: (Please note that the current is in milliamperes) mA Similarly the current across R2 is: 12 = mA And the current across R3 is: 13 = mA The current read by the ammeter is the sum of the currents across each resistor: 1=1+/2+13 | = mA Alternatively, we can solve for the current across the ammeter by first calculating the equivalent resistance and then applying Ohm's law. The equivalent resistance of the three resistors in parallel is: 1/Reg =1/R1+1/R2 +1/ Thus, Reg Ω

A wire of one metal has twice the resistivity, twice the diameter and twice the length of a wire of another metal. What is the ratio of the resistances of the two wires? Give a numberical value and explain your reasoning.

In an experiment, an initially uncharged capacitor of capacitance C=62µF (microFarad) is connected to a resistor R and a battery of voltage V as shown in the figure below: V(V) VT After the switch is closed at t=0, the voltage across the capacitor is measured and plotted as shown in the figure below. Using the data provided in the curve, determine the value of resistance R. Express your answer in units of using zero decimal places. 13- 12- 11 10- 9 8- 7- 6 5 4- 3 2 1- 0- 0 P 0.1 66****** 0.2 0.3 R www 0.4 0.5 C 0.6 t (s) 0.7 0.8 0.9 1 1.1 1.2 4

E1, r E2, r 26 N A В C I