1. Analyzing the External Circuit (Left Side)

We first analyze the known components connected to the left terminals (A and B) of the black box to determine the properties of the non-ideal meters and the voltage/current state at the terminals.

Applying Kirchhoff’s Current Law (KCL):
Let’s observe the junction between the two ammeters and the voltmeter.

• Current $I_{A2}$ (reading $20\text{ mA}$) is on the right, connected to the black box.
• Current $I_{A1}$ (reading $10\text{ mA}$) is on the left, connected to the $3.0\text{ V}$ battery.

For the currents to be consistent with the circuit topology (a source on the far left and a load in the box), the current must flow out of the black box at terminal A.
At the junction: $$ I_{from\_box} = I_{voltmeter} + I_{to\_battery} $$ $$ 20\text{ mA} = I_V + 10\text{ mA} $$ $$ \Rightarrow I_V = 10\text{ mA} $$

Calculating Meter Resistances:

• Voltmeter Resistance ($R_V$): The voltmeter reads $3.5\text{ V}$ with a current of $10\text{ mA}$. $$ R_V = \frac{V_1}{I_V} = \frac{3.5\text{ V}}{10\text{ mA}} = 350\,\Omega $$
• Ammeter Resistance ($R_A$): Consider the left loop containing the battery, Ammeter 1, and Voltmeter 1.
  The potential at the junction is $3.5\text{ V}$. The battery potential is $3.0\text{ V}$. The current $10\text{ mA}$ flows into the battery (charging it).
  Voltage drop across Ammeter 1: $\Delta V_{A1} = 3.5\text{ V} – 3.0\text{ V} = 0.5\text{ V}$. $$ R_A = \frac{\Delta V_{A1}}{I_{A1}} = \frac{0.5\text{ V}}{10\text{ mA}} = 50\,\Omega $$

2. Determining Conditions at the Black Box Terminals

Left Terminals (A-B):

The voltage at terminal A ($V_A$) is higher than the junction voltage because current flows from A to the junction through Ammeter 2. $$ V_A = V_{junction} + I_{A2} R_A $$ $$ V_A = 3.5\text{ V} + (20\text{ mA})(50\,\Omega) $$ $$ V_A = 3.5\text{ V} + 1.0\text{ V} = 4.5\text{ V} $$ So, the black box maintains a terminal voltage of 4.5 V at A relative to B while supplying 20 mA.

Right Terminals (C-D):

The output is connected to a voltmeter ($V_2$) reading $3.0\text{ V}$. Since the meters are identical, the load resistance here is $R_V = 350\,\Omega$.
So, at terminals C-D, the voltage drops to 3.0 V when loaded by $350\,\Omega$.

3. Designing the Simplest Internal Circuit

We need a circuit that:

1. Provides a stiff $4.5\text{ V}$ at terminals A-B (behaving like an ideal voltage source).
2. Provides $3.0\text{ V}$ at terminals C-D when loaded with $350\,\Omega$.