Answer to Question #194685 in Electricity and Magnetism for Mia

Question

A 9 V battery is connected in series with a 10 resistor. What is the current in the circuit, to the nearest tenth of an ampere?

Answer 1

The voltage in the circuit is U = 9 V and the total resistance is (if we consider the battery as an ideal battery, so the resistance of it is 0) R = 10 Ohm. According to the Ohm's law, the current I is $I=\dfrac{U} {R} = \dfrac{9\,\mathrm{V}}{10\,\mathrm{Ohm}} =0.9\,\mathrm{A}.$

Well, it is unclear from the statement of the task if the resistance of the resistor is 10 Ohm (and Ohm is omitted), or the resistance is 1 Ohm (and symbol $\Omega$ became O). If it is the last case, the current will be 10 times bigger, or 9 A.