In the association of resistors on the side,

an ideal voltage source of 10 V has been connected

between points A and B. Determine:

(a) The equivalent resistance of the circuit.

(b) The total electrical current in the circuit.

(c) The potential difference across the 7 ohm resistor
terminals.

(d) The electric current through the 4 ohm resistor.

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3,9 are in geriepR_{1}=3+9=12 \Omega4, R_{1} are in parallel10R_{2}=\frac{(4)(12)}{4+12}=3 \Omega\begin{aligned}R_{\text {eq }} & =5 \|(3+7) \\& =\frac{(5)(10)}{5+10} \\R_{e q} & =3.333 \Omega \\I_{T} & =\frac{10}{R_{e q}}=\frac{10}{3.333} \\I_{T} & =3 \mathrm{~A}\end{aligned} 10\begin{array}{l}79 \\\end{array}Apply mesh analysis for loop is\begin{aligned}-10+5\left(i_{1}-i_{2}\right) & =0 \\5 i_{1}-5 i_{2} & =10 \rightarrow \text { (1) }\end{aligned}Apply mesh analysis for loop i2\begin{array}{c}5\left(i_{2}-i_{1}\right)+4\left(i_{2}-i_{3}\right)+7 i_{2}=0 \\16 i_{2}-5 i_{1}-4 i_{3}=0 \rightarrow \text { (2) }\end{array}Apply mesh analysis for loop i_{3}\begin{array}{c}4\left(i_{3}-i_{2}\right)+3 i_{3}+9 i_{3}=0 \\16 i_{3}-4 i_{2}=0 \rightarrow(3\end{array}solving eq (1), (2) & (3)\begin{array}{l}i_{1}=3 \mathrm{~A} \\i_{2}=1 \mathrm{~A} \\i_{3}=0.25 \mathrm{~A}\end{array}v_{7}=7 vCurrent through 4 \Omega \quad i_{4}=i_{2}-i_{3}=1-0.25i_{4}=0.75 \mathrm{~A} ...