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We can model a certain battery as a voltage source in series with a resistance. The open-circuit voltage of the battery is 10 V . When a 120-Ω resistor is placed across the terminals of the battery, the voltage drops to 7 V. Determine the internal resistance of the battery.

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Answer:

51.4 Ohms

Explanation:

By applying voltage division rule

[tex]V_f=v_i\times \frac {R_l}{R_l+R_m}[/tex] where v is voltage, subscripts i and f represnt initial and final, R is resistance, m is internal and l is external.Substituting 7V for final voltage, 10V for initial voltage and the external resistance as 120 Ohms then

[tex]7=10*\frac {120}{120+R_m}\\7R_m+840=1200\\R_m={1200-840}{7}=51.428571\approx 51.4 Ohms[/tex]

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