question archive An ammeter is connected in series to a battery of voltage Vb and a resistor of unknown resistance Ru (Figure 1)
An ammeter is connected in series to a battery of voltage Vb and a resistor of unknown resistance Ru (Figure 1)
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An ammeter is connected in series to a battery of voltage Vb and a resistor of unknown resistance Ru (Figure 1) . The ammeter reads a current Io. Next, a resistor of unknown resistance Rr is connected in series to the ammeter, and the ammeter's reading drops to I1. Finally, a second resistor, also of resistance Rr, is connected in series as well. Now the ammeter reads I2.
If I1/I0=4/5, find I2/I0.
Express the ratio I2/I0 numerically.
I2/I0 =
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If I1 / I0 = 4/5, find I2 / I0.
Express the ratio I2 / I0 numerically.
Since I1 / I0 = 4/5, we can say that I1 = 4 and I0 = 5. We also know that V = IR and I = V/R. Say that Ru = 1. That means that I0 = 5 = V / 1, so V = 5. So for I1:
V = IR
The resistors are in series, so:
5 = 4 * (Ru + Rr)
5 = 4 * (1 + Rr)
Rr = 0.25
We can apply this to I2 (again, the resistors are in series, so we add them):
V = I2 * R
V = I2 * (Rr + Rr + Ru)
5 = I2 (1.5)
I2 = 3.333
So I2 / I0 = 3.333 / 5 = 0.667
