Suppose you measure the terminal voltage of a $1.585-\mathrm{V}$ alkaline cell having an internal resistance of 0.100$\Omega$ by placing a 1.00 -k $\Omega$ voltmeter across its terminals. (See Figure 21.57.) (a) What current flows? (b) Find the terminal voltage. (c) To see how close the measured terminal voltage is to the emf, calculate their ratio.
Added by Chase W.
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585 \, \text{V}\) and \(R_{\text{eq}} = R_{\text{internal}} + R_{\text{voltmeter}} = 0.1 \, \Omega + 1000 \, \Omega = 1000.1 \, \Omega\). Therefore, \(I = \frac{1.585}{1000.1} = 0.001584 \, \text{A}\). Show more…
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Suppose you measure the terminal voltage of a $1.585-\mathrm{V}$ alkaline cell having an internal resistance of $0.100 \Omega$ by placing a 1.00 - $\mathrm{k} \Omega$ voltmeter across its terminals. (See Figure 21.54.) (a) What current flows? (b) Find the terminal voltage. (c) To see how close the measured terminal voltage is to the emf, calculate their ratio.
Suppose you measure the terminal voltage of a 1.585-V alkaline cell having an internal resistance of $0.100 \Omega$ by placing a $1.00-\mathrm{k} \Omega$ voltmeter across its terminals (see below). (a) What current flows? (b) Find the terminal voltage. (c) To see how close the measured terminal voltage is to the emf, calculate their ratio.
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