Consider the circuit, in which three resistors of resistances $R, 2 R$, and $R$ are connected to batteries of voltages $V$ and $2 V$, as shown in the figure below. If $R=45 \Omega$ and $V=15 \mathrm{~V}$, determine $I$. Express your answer in mA using zero decimal places. Please note that depending on the direction of the current, your answer may be negative. If that is the case for you, please do not forget the minus sign!
Answer
\boxed{2333 \text{ mA}}
Step 1 ：Apply Kirchhoff's Voltage Law (KVL) to the two loops in the circuit: Loop 1: \(V - I_1 R - I_2 2R = 0\) and Loop 2: \(2V - I_3 R - I_1 R = 0\)
Step 2 ：Apply Kirchhoff's Current Law (KCL) at the junction between the resistors: \(I_1 - I_2 = I_3\)
Step 3 ：Plug in the given values for R and V: \(R = 45\) and \(V = 15\)
Step 4 ：Solve the system of equations for I1, I2, and I3
Step 5 ：\(I = I_1 - I_2\)
Step 6 ：\boxed{2333 \text{ mA}}
