The circuit model of a battery charger and battery is drawn in Fig. 2.62. The variable resistance $\mathrm{R}$ controls the charging current.

Determine $R$ for

(a) Charging current of $5 \mathrm{~A}$

(b) Power delivered to battery of $12.5 \mathrm{~W}$

(c) Voltage at battery terminals of $13 \mathrm{~V}$

Question

The circuit model of a battery charger and battery is drawn in Fig. 2.62. The variable resistance [latex]\mathrm{R}[/latex] controls the charging current.

Determine [latex]R[/latex] for

(a) Charging current of [latex]5 \mathrm{~A}[/latex]

(b) Power delivered to battery of [latex]12.5 \mathrm{~W}[/latex]

(c) Voltage at battery terminals of [latex]13 \mathrm{~V}[/latex]

Accepted Answer

(a) By KVL

[latex] \begin{aligned} -15+(0.02+R+0.035) I+12.5 & =0 \ (R+0.055) I=15-12.5 & =2.5 \ I & =5 \mathrm{~A} \end{aligned} [/latex]

Then

[latex] \begin{aligned} & 5 R+0.275 =2.5 \ \mathrm{Or} \qquad & R =0.445 \Omega \end{aligned} [/latex]

(b) Power delivered at battery terminals

[latex] \begin{gathered} \mathrm{P}=12.5 I+0.035 I^{2}=25 \mathrm{~W} \ 0.035 I^{2}+12.5 I-25=0 \ I=\frac{-12.5 \pm \sqrt{(12.5)^{2}+4 imes 25 imes 0.035}}{2 imes 0.035}=1.99 \mathrm{~A} \end{gathered} [/latex]

Solving we get

[latex]I=1.99[/latex] A, negative value is rejected.

(c) Voltage at battery terminals [latex]=13 \mathrm{~V}[/latex]

[latex]I=\frac{13-12.5}{0.035}=14.28 \mathrm{~A}[/latex]

KVL round the loop

[latex]15-(0.02+R) imes 14.28=13[/latex]

Solving we get

[latex]R=0.12 \Omega[/latex]

Question 2.ASP.33: The circuit model of a battery charger and battery is drawn ...

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