Suppose that the resistance between the walls of a biological cell is 5.0 times

Question

Suppose that the resistance between the walls of a biological cell is 5.0 times 10^9 Ohm. (a) What is the current when the potential difference between the walls is 75 mV? (b) If the current is composed of Na^+ ions (q = + e), how many such ions flow in 0.50 s? A resistor is connected across the terminals of a 9.0-V battery, which delivers 1.1 times 10^5 J of energy to the resistor in six hours. What is the resistance of the resistor? The current in a 47-Ohm resistor is 0.12 A. This resistor is in series with a 28-Ohm resistor, and the series combination is connected across a battery. What is the battery voltage? A 16-Ohm loudspeaker, an 8.0-Ohm loudspeaker, and a 4.0-Ohm loudspeaker are connected in parallel across the terminals of an amplifier. Determine the equivalent resistance of the three speakers, assuming that they all behave as resistors. Find the equivalent resistance between points A and B in the drawing.

Explanation / Answer

5) Resistance = 5*10^9 ohm

I = V/R = 0.075 / 5*10^9 = 1.5*10^-11 A

I = q/t

q = 1.5*10^-11*0.5 = 7.5*10^-12 C

number of electrons flowing = 7.5*10^-12 / (1.6*10^-19) = 4.6875 *10^7

7) Power = V^2/R

1.1*10^5 / 6*60*60 = 9^2/R

R = 15.9 ohm

45) R (total) =R1 + R2 = 47 + 28 = 75 ohm

V = IR = 0.12*75 = 9 V

54 ) 1/R = 1/R1 + 1/R2+1/R3

1/R = 1/16 + 1/8 + 1/4

R = 2.28 ohm

64 ) 1 , 2 and 3 ohm are in series

R1 = 1 + 2 + 3 = 6 ohm

3 and 6 ohm are in parallel

R2 = 3*6 / (3 +6 ) =2 ohm

2 and 4 ohm are in parallel

R3 = 2*4 /(2+4) = 1.33 ohm

1.33 and 2 ohm are in series

R4 = 1.33 + 2 = 3.33 ohm

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