You want to make a current source from a battery (V) and resistor (R) so that the current varies by less than 10% as the load resistance (R) changes from 200 to 500N. The current must be at least 1.5 mA. For V = 10 Volts, find a resistance R which will meet these criteria. 1) R I V R
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- Ten resistors, each having a resistance of 120 Ohms are arranged as follows. Find the totalresistance.The emf and the internal resistance of a battery are as shown in the figure below. If a current of 3.2 A is drawn from the battery when a resistor R is connected across the terminals ab of the battery, what is the power dissipated by the resistor R? 95.0 V 5.0 2The image shows a circuit diagram that includes a battery with an emf of 10.0 V, some resistors, and an ammeter. If the battery is ideal, find the equivalent resistance of the circuit and the current through the ammeter. 35.0 2 a) Req = 10.0 V b) Iummeter = c) If the battery has an internal resistance of 1.30 Q, find the current through the ammeter. 42.0 2 A 7.00 2 U O'DI
- Choose the letter of correct answer. Resistance 1 Ω, 2 Ω, and 3 Ω are connected in the form of triangle. If a cell of e.m.f 1.5 V and negligible internal resistance is connected across 3 Ω, the current through this resistance will be: a. 0.25 A b. 0.5 A c. 1.5 A d. 2.5 AA. Ammeter-Voltmeter Methods DATA TABLE 1 Purpose: To measure resistance values. 29.3 Voltmeter resistance R Accepted value of R DATA TABLE 2 24.6 Purpose: To measure resistance values. Rheostat Current I setting Rh ( ) Voltage V ( ) Resistance R ( ) Rheostat setting Rh Current I Voltage V R' = V/I ( ) ( ) () 1 ,011) A 1.760 1 .0117A 1.758 2 .0112 A 1.669 2 .0113 A 1.700 3 . 0106 A 1.592 3 .0107A 1.611 Average R Average R' Calculations (show work) Percent error of R Percent error of R' Ammeter resistance R 3The potential difference across a resting neuron in the human body is about 63.0 mV and carries a current of about 0.215 mA. How much power does the neuron release? Need Help? Read It
- A battery with an emf of 24 V is connected to 6.0 Q resistor. As a result current of 3.0 A exists in the resistor. The internal resistance of the battery is: Select one: O a. 6.0 0 O b. 1.0 2 O C. 2.0 Q O d. 4.0 2potentiometer has uniform potential gradient. The specific resistance of the material of the potentiometer wire is 10-9Ωm and the current passing through it is 0.5A and the cross sectional area of the wire is 10-8m2. Calculate the potential gradient along the potentiometer wire. a) 0.5 × 10-9 V/m b) 0.5 × 10-8 V/m c) 0.5 × 10-1 V/m d) 0.5 × 10-15 V/mConsider the circuit shown in the figure below. (Assume R₁ = 10.5 and R₂ = 3.50 2.) b 25.0 V + T R₁ R₁ a R₂ R₂ (b) Find the current in the 20.0- resistor. A -20.0 Ω (a) Find the potential difference between points a and b. V
- a)(QUESTION 2) Determine the length (in mm) of a gold conductor that has a resistance of 0.018 Ω and a cross-sectional area of 3.4 x 10-9 m2. The resistivity of gold is 2.2 x 10-8 Ω-m b) An aluminum conductor with the same length and area as in Question 2 above would have a higher resistance than the gold conductor. The resistivity of aluminum is 2.6 x 10-8 Ω-m. True or False?A resistor of 1.8 kΩ has a potential of 4.0 x 10^2 V. a) What is the current through the resistor? b) What is the power of the resistor? c) How much thermal energy is released after 1 hour? Remember to include the following when answering these questions: What did you get? (your answer with the correct units and supporting work) How did you get that? (the equation you chose to use to get your answer) Why did you use that? (the concept that supports the use of the equation that you chose to use)Part 1 Refer to "Real Batteries" in your textbook in addition to the lab manual. (1) A battery has an emf of 9.0 V. After connecting a light bulb, you measure the terminal voltage of the battery between points a and b to be 8.81 V and the current in the light bulb to be 752 mA. What is the internal resistance of the battery? (2) What is the resistance of the light bulb? a R b