The voltage, V, in a simple electrical circuit is slowly decreasing as the battery wears out. The resistance, R is slowly increasing as the resistor heats up. Use Ohm's Law, V = IR to find the rate at which the current, J, is dropping at the moment when -0.002 Y, and dR dV R = 100 N. I = 0.12 A. dt = 0.025 4. dt Note: voltage is measured in volts (V), resistance is measured in ohms (), and current is measured in amperes (A); 1 V = (1 A) · (1 N).

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The voltage, V, in a simple electrical circuit is slowly decreasing as the battery wears
out. The resistance, R, is slowly increasing as the resistor heats up. Use Ohm's Law,
V = IR to find the rate at which the current, Į, is dropping at the moment when
= -0.002
R = 100 N. I = 0.12 A.
dV
dt
and
dt
dR
0.025 2.
Note: voltage is measured in volts (V), resistance is measured in ohms (2), and
current is measured in amperes (A): 1 V = (1 A) · (1 N).
Transcribed Image Text:The voltage, V, in a simple electrical circuit is slowly decreasing as the battery wears out. The resistance, R, is slowly increasing as the resistor heats up. Use Ohm's Law, V = IR to find the rate at which the current, Į, is dropping at the moment when = -0.002 R = 100 N. I = 0.12 A. dV dt and dt dR 0.025 2. Note: voltage is measured in volts (V), resistance is measured in ohms (2), and current is measured in amperes (A): 1 V = (1 A) · (1 N).
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