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3 years ago

Humberto builds two circuits using identical components.

2 answers:

6 0

Humberto should expect to see that all bulbs in circuit 1 will shine more dimly than the original bulbs, while all bulbs in circuit 2 will have the same brightness as the original bulbs.

B) All bulbs in circuit 1 will shine more dimly than the original bulbs, while all bulbs in circuit 2 will have the same brightness as the original bulbs.

Bumek [7]3 years ago

5 0

Answer:

B) All bulbs in circuit 1 will shine more dimly than the original bulbs, while all bulbs in circuit 2 will have the same brightness as the original bulbs.

Explanation:

When we connect more bulbs in circuit 1 where all are connected in series then the net resistance of the circuit will increase and voltage across each bulb will decrease.

So as per the formula

$P = \frac{V^2}{R}$

so here since voltage decreases across each bulb so we can say that power across each bulb will decrease and bulb gets more dimmy then initially connected bulbs

When we connect bulbs in parallel then voltage across each bulb will remain same and then the brightness of each bulb will remain constant.

So correct answer will be

B) All bulbs in circuit 1 will shine more dimly than the original bulbs, while all bulbs in circuit 2 will have the same brightness as the original bulbs.

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A voltaic cell with an aqueous electrolyte is based on the reaction between Cd2+(aq) and Mg(s), producing Cd(s) and Mg2+(aq). Wr

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Answer : The balanced two-half reactions will be,

Oxidation half reaction (anode) : $Mg(s)\rightarrow Mg^{2+}(aq)+2e^-$

Reduction half reaction (cathode) : $Cd^{2+}(aq)+2e^-\rightarrow Cd(s)$

Thus the overall reaction will be,

$Mg(s)+Cd^{2+}(aq)\rightarrow Mg^{2+}(aq)+Cd(s)$

Explanation :

Voltaic cell : It is defined as a device which is used for the conversion of the chemical energy produces in a redox reaction into the electrical energy. It is also known as the galvanic cell or electrochemical cell.

The given redox reaction occurs between the magnesium and cadmium.

In the voltaic cell, the oxidation occurs at an anode which is a negative electrode and the reduction occurs at the cathode which is a positive electrode.

The balanced two-half reactions will be,

Oxidation half reaction (anode) : $Mg(s)\rightarrow Mg^{2+}(aq)+2e^-$

Reduction half reaction (cathode) : $Cd^{2+}(aq)+2e^-\rightarrow Cd(s)$

Thus the overall reaction will be,

$Mg(s)+Cd^{2+}(aq)\rightarrow Mg^{2+}(aq)+Cd(s)$

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