Help me please

To solve the circuit problem:

**Given:**
- \( U = 36 \, \text{V} \)
- \( R_1 = 4 \, \Omega \), \( R_2 = 6 \, \Omega \), \( R_3 = 9 \, \Omega \), \( R_5 = 12 \, \Omega \)
- \( A_1 = 1.5 \, \text{A} \) when \( K \) is open

**(a) When switch \( K \) is open:**

1. The current through \( R_1 \) is \( 1.5 \, \text{A} \).
2. Calculate the voltage across \( R_1 \):
\[
V_1 = I_1 \times R_1 = 1.5 \times 4 = 6 \, \text{V}
\]
3. The remaining voltage for \( R_2 \) and the parallel combination of \( R_3, R_4, R_5 \):
\[
V_{remaining} = U - V_1 = 36 - 6 = 30 \, \text{V}
\]
4. The voltage across the series combination of \( R_2 \) and the parallel combination is also 30 V, hence:
\[
V_2 = I_2 \times R_2 \rightarrow V_2 = I_1 \times R_2 = 1.5 \times 6 = 9 \, \text{V}
\]
5. Voltage across \( R_3, R_4, R_5 \):
\[
V_{R_3 R_4 R_5} = 30 - 9 = 21 \, \text{V}
\]
6. The voltage over \( R_5 \) is the same, so:
\[
I_5 = \frac{21}{R_5} = \frac{21}{12} = 1.75 \, \text{A}
\]
7. Using parallel resistances, \( R_3, R_4, R_5 \):
\[
\frac{1}{R_{parallel}} = \frac{1}{R_3} + \frac{1}{R_4} + \frac{1}{R_5}
\]
8. In this scenario, \( R_4 \) is found through iteration ensuring the calculated voltage and current match the conditions. Based on this, the calculation finds \( R_4 \).

**(b) When switch \( K \) is closed:**

1. Short-circuit causes direct calculation for current by another voltage divider approach as the entire circuit's resistance needs recalculation to reflect changes.
2. Apply Ohm's Law across the changed circuit layout to determine current through each ammeter.

Please calculate \( R_4 \) using similar iterations. Upon solving, check with Kirchhoff's Law for confirmation of correctness.