Three resistors are connected as shown in the diagram below. Two resistors are in parallel with a 6 Ω resistor. The current in the ammeter is 6 A.

Question

2.1 The current in ammeter 2 is Answer 1 Question 7 A.
2.2 The current through the 4 Ω is Answer 2 Question 7 A.
2.3 The equivalent resistance of the resistors is Answer 3 Question 7 Ω.

Answer 1

To solve the problem, we'll first examine the configuration of the resistors based on the information provided.

Given:

• The total current (I_total) in the circuit is 6 A.
• We have a 6 Ω resistor and two other resistors that are in parallel. From the context, it appears the other resistors are 4 Ω and an unknown resistor.

2.1 Current in ammeter 2: Since the problem might imply a parallel connection between the 6 Ω resistor and the other two resistors, the current splits at the junction.

For resistors in parallel:

• Let I_6 be the current through the 6 Ω resistor.
• Let I_4 be the current through the 4 Ω resistor.
• Let I_unknown be the current through the unknown resistor.

According to Kirchhoff's current law (KCL): \[ I_{\text{total}} = I_6 + I_4 + I_{\text{unknown}} \]

However, without detailed configurations, we cannot definitively find the currents without knowing the total resistance of the parallel network.

Assuming the remaining two resistors (4 Ω and unknown) are in series or parallel as suggested, generally:

Using Ohm's Law: \[ V = I \cdot R \]

In a parallel circuit: \[ \frac{1}{R_{\text{eq}}} = \frac{1}{R_1} + \frac{1}{R_2} \]

We could estimate the current through the 6 Ω resistor using Ohm's law: \[ I_6 = V/R_6 \]

2.2 To find the current through the 4 Ω resistor, we would need to calculate the voltage across the 6 Ω resistor first:

The voltage across the 6 Ω resistor (using Ohm's Law): \[ V = I_{\text{total}} \cdot R = 6 \cdot 6 = 36 V \]

This is the same voltage across the 4 Ω resistor since they are in parallel. Now, we can find the current through the 4 Ω resistor:

\[ I_4 = \frac{V}{R} = \frac{36 V}{4 Ω} = 9 A \]

2.3 For the equivalent resistance of the circuit: If we have a parallel configuration:

\[ R_{\text{eq}} = \frac{1}{\frac{1}{R_1} + \frac{1}{R_2}} + R_3 \]

Without needing to find an unknown resistor: \[ \frac{1}{R_{\text{eq}}} = \frac{1}{6} + \frac{1}{4} \] \[ \frac{1}{R_{\text{eq}}} = \frac{2}{12} + \frac{3}{12} = \frac{5}{12} \] \[ R_{\text{eq}} = \frac{12}{5} Ω = 2.4 Ω \]

Thus,

• Answer 2.1 (current through ammeter 2): 6 A (assuming all current passed through.)
• Answer 2.2 (current through 4 Ω resistor): 9 A
• Answer 2.3 (equivalent resistance): 2.4 Ω

Please clarify any component connections to assist with further specific answers!