An ammeter is connected in series to a battery of voltage V_b and a resistor of unknown resistance R_u The ammeter reads a current I_0. Next, a resistor of unknown resistance R_r is connected in series to the ammeter, and the ammeter's reading drops to I_1. Finally, a second resistor, also of resistance R_r, is connected in series as well. Now the ammeter reads I_2.

If I_1/I_0= 4/5, find I_2/I_0.

express numerically. I hate when they ask us this stuff without numbers. I could do it if I had the resistances.

Its not so difficult...

Io=Vb/Ru I1=Vb/(Ru+Rr)

so IoRu=I1(Ru+Rr)

then 4/5=Ru/(Ru+Rr) or solving...
Ru=4Rr. or Rr=1/4 Ru. If you add a second Ru, then Rtotal=Ru+ .5Ru or 3/2 Ru

I2/Io= Ru/(1.5Ru)= 2/3

check my thinking

wow your right i feel stupid. i thought the other resistor added was different. now it makes sense thanks.

To find the ratio I_2/I_0, we can use the concept of current division in a series circuit.

Let's denote the resistance of the ammeter as R_a and the resistance of the battery as R_b. Since the ammeter is connected in series with the battery and the unknown resistor R_u, the total resistance in this circuit is R_t = R_a + R_b + R_u.

From the given information, we have:
I_0 = V_b / R_t (Equation 1)

When the resistor R_r is connected in series with the ammeter, the total resistance becomes R_t' = R_a + R_b + R_u + R_r. Using current division, we can express I_1 in terms of I_0:

I_1 = I_0 * ( R_t / R_t' ) (Equation 2)

And when the second resistor of resistance R_r is also connected in series, the total resistance becomes R_t'' = R_a + R_b + R_u + 2R_r. Again, using current division, we can express I_2 in terms of I_0:

I_2 = I_0 * ( R_t / R_t'' ) (Equation 3)

To find I_2/I_0, we need to simplify Equation 3 by substituting the value of R_t'' using R_t'' = R_t + R_r:

I_2 = I_0 * ( R_t / ( R_t + R_r ) ) (Equation 4)

Dividing Equation 4 by Equation 1 will give us the desired ratio:

I_2/I_0 = ( I_0 * ( R_t / ( R_t + R_r ) ) ) / I_0 = R_t / ( R_t + R_r )

Since we don't have numerical values for the resistances, we cannot determine I_2/I_0 explicitly. However, we can say that I_2/I_0 depends on the values of R_t and R_r.

To find I_2/I_0, we can use the relationship between the currents I_0, I_1, and I_2.

Let's assume the resistance of the ammeter is negligible compared to the other resistances involved.

First, let's analyze the circuit with just the battery (V_b) and the resistor R_u.

In this case, the current passing through the ammeter (I_0) is given by Ohm's Law: I_0 = V_b / (R_u + 0) = V_b / R_u.

Next, when the resistor R_r is connected in series with the ammeter, the current passing through the circuit decreases. We know that I_1/I_0 = 4/5, so I_1 = (4/5) * I_0.

Now, when a second resistor R_r is connected in series as well, the total resistance in the circuit increases and therefore, the total current passing through the circuit decreases. Let's call this current I_2.

To find I_2/I_0, we need to consider the equivalent resistance in the circuit when both resistors R_r are added.

Since the resistors are connected in series, the equivalent resistance (R_eq) is the sum of the resistances: R_eq = 2 * R_r.

Now, using Ohm's Law again, the current passing through the circuit with two resistors is given by: I_2 = V_b / (R_u + R_eq) = V_b / (R_u + 2 * R_r).

To find I_2/I_0, we divide I_2 by I_0:

I_2/I_0 = (V_b / (R_u + 2 * R_r)) / (V_b / R_u) = R_u / (R_u + 2 * R_r).

Now we have the expression for I_2/I_0 in terms of R_u and R_r.

Unfortunately, without specific values for R_u and R_r, we cannot provide a numerical answer. However, you can use the given expression and substitute the specific values of R_u and R_r when given to calculate the ratio I_2/I_0.